Math 601-602 — 2009

Summary

Linear Algebra

21 January

Matrices.  There are several pieces of notation that we will use. Real numbers are always denoted by ℝ, complex by ℂ, and m×n matrices with real or complex entries by ℝm×n and ℂm×n, respectively. We also discussed matrix algebra: equality of matrices, addition of matrices, multiplication by a scalar, ROW×COLUMN (dot or inner product), matrix multiplication, COLUMN×ROW (outer product), and the transpose and adjoint (conjugate transpose) of a matrix. See the article in Wikipedia on Matrix Multiplication. (The online text covers it, but not in the order that we need.)

The algebra corresponding to these operations is the same as that for ordinary numbers, except for these two things:

  1. AB ≠ BA
  2. AB = 0 does NOT imply that either A or B is 0. For example, if A = B =
     1    1
    -1   -1
    
    then AB = 0.
Systems of linear equations.  It's useful to write systems in matrix form. While we looked at several examples, including the general form of a system of linear equations, it's useful just to look at a simple, special case. The system
3x + 2y − 4z = 1
5x + 6y − 2z = 3
has the matrix form Ax = b, where x = (x y)T, b = (1 3)T, and A =
3 2 -4
5 6 -2
A is called the coefficient matrix, x is the vector of unknowns, and b is the inhomogeneous term (or just RHS - right hand side). The first chapter of the text by Hefferon gives many more examples.

23 January

Classifying systems of linear equations.  Linear systems are classified in several ways. The first is whether b = 0. If it is, the system is called homogeneous; if b0, it's inhomogeneous. The second way is according to the number of equations, m, and the number of unknowns, n:

Underdetermined   Determined   Overdetermined
m < n m = n m > n

As we shall see, underdetermined systems always have an infinite number of solutions. Determined systems usually, but not always, have a single solution. Overdetermined systems are usually a problem and have to be carefully treated. Frequently, they have no solutions.

Solving systems of linear equations.  There is a systematic method for solving systems of linear equations, the Gauss-Jordan method. We discussed how this method works in connection with manipulating the equations in a system. If the system is given in terms of matrices, Ax = b, then we can simplify the method by using the augmented matrix, [A|b]. For example, the system below

3x + 2y − 4z = 1
5x + 6y − 2z = 3
has the augmented matrix
3 2 -4 | 1
5 6 -2 | 3
The advantage is that we will only have to deal with the rows of [A|b], rather than carrying the the unknown variables in equations. While this makes by-hand calculations easier, it is invaluable for computer calculations.

The three row operations below replace corresponding manipulations with equations.

  1. Multiply a row by a nonzero constant.
  2. Replace a given row Rj by the sum Rj+cRk, where k≠j and c is any scalar.
  3. Interchange two rows.
These three operations are systematically used in a process called row reduction. We did several examples in class. The system below
x + y + z = 1
3x − y + z = 1
has the augmented matrix
1  1 1 | 1
3 -1 1 | 1
This matrix is row-equivalent to the matrix
1  0 ½ | ½
0  1 ½ | ½
which is the augmented matrix for the system
x + z/2 = 1/2
y + z/2 = 1/2
It is customary to replace z by a parameter, say t. The we write the system as
x = 1/2 - t/2
y = 1/2 - t/2
z = t
In column vector form, this is x = t(-1/2 -1/2 1)T + (1/2 1/2 0)T

26 January

Row reduction and rank of a matrix.  These topics are discussed in the Notes on Row Reduction and the notes on The Rank of a Matrix. We went over these in detail in class.

Inverse of a square matrix.  Consider the matrix A =

1 -1
2 -1
Suppose that we want to solve Ax = b for any choice of b — for instance, b = (5 7)T. To do this, we can solve two systems of equations, Ac1 = (1 0)T and Ac2 = (0 1)T. Using row reduction, we see that c1 = (-1 -2)T and c2 = (1 1)T. Notice that
5Ac1 + 7Ac2 = (5 7)T.
However, we also have that
5Ac1 + 7Ac2 = A(5c1 + 7c2).
Thus the column x = 5c1 + 7c2 solves Ax = (5 7)T. In fact, using the "basic matrix trick" (cf. The Rank of a Matrix), x = [c1 c2](5 7)T = (2 -3)T. The matrix A-1 = [c1 c2] =
-1 1
-2 1
is called the inverse of A.

There are two points to made here. The first is that x = [c1 c2]b solves Ax = b in general. The second is that solving for c1 and c2 can be done simultaneously, using only one set of row operations.


28 January

Inverse of a square matrix (continued).  Let A be an n×n matrix. We say the A has an inverse B if and only if AB = BA = I, where I is the n×n identity. As we pointed out above, not all square matrices have inverses. If a matrix A does have an inverse B, then that inverse is unique; we write A-1 := B and say A is invertible or nonsingular.

Algorithm for finding the inverse.  Finding the inverse of a matrix is equivalent to solving n systems of equations whose solutions are the columns of the inverse of the matrix, if that inverse exists. In fact there is a very simple algorithm that either gives the inverse of a matrix A or shows that one doesn't exist.

Algorithm
  1. Form the augmented matrix [A|I].
  2. Row reduce [A|I].
Here are two examples of the algorithm.

Example 1.  Let A =

1  1  1       
2  0 -1       
1 -1  1    
The augmented matrix is [A|I] =
1  1  1  | 1 0 0  
2  0 -1  | 0 1 0
1 -1  1  | 0 0 1
Row reducing [A|I] results in the matrix
1  0  0  | 1/6   1/3  1/6  
0  1  0  | 1/2    0  -1/2    
0  0  1  | 1/3  -1/3  1/3  
Thus, A-1 =
1/6   1/3  1/6
1/2    0  -1/2
1/3  -1/3  1/3

Example 2.  Let A =

 1   1   2       
 2   0   2       
-2  -1  -3    
The augmented matrix is [A|I] =
 1   1   2  | 1 0 0  
 2   0   2  | 0 1 0
-2  -1  -3  | 0 0 1
Row reducing [A|I] results in the matrix
1  0  0  | 1   0   1/2  
0  1  1  | 1   0   -1   
0  0  0  | 1  1/2   1  
Since the 3×3 matrix on the left has a row of zeros, the matrix A is singular.

The theorem below summarizes conditions for a matrix to have an inverse. We assume that A is an n×n matrix with real or complex entries. The last condition on the list below involves the determinant, which we will discuss later. To make the list complete, we include it here.

Theorem 1/28.1. The matrix A is invertible (nonsingular) if and only if any one of the following equivalent conditions hold.
  1. [A|I] is row equivalent to [I|B].
  2. Rank(A) = n.
  3. The columns of A are linearly independent.
  4. The homogeneous equation Ax = 0 has only x = 0 as a solution.
  5. det(A) ≠ 0.
2×2 inverse.  The 2×2 case comes up often enough and is simple enough that it is worth doing in general. Consider the 2×2 matrix
a b
c d
With a little algebra, one can show that A is invertible if and only if det(A) = ad - bc ≠ 0. When this happens, A-1 = det(A)-1×
 d  -b
-c   a

30 January

Determinants.  The lecture for today is from my notes on Determinants. I want to mention that many of you have seen and used the "arrow method" for finding 3×3 determinants. This method is simply the permutation definition, with a systematic way of keeping track of the signs of the six permutations used in computing a 3×3 matrix. It does not work in higher dimensions.


2 February

Methods for finding determinants.  We discussed several methods for finding determinants: (1) the method of cofactors and (2) the method of using row (or column) operations to put the determinant in triangular form. We started Cramer's. See my notes on Determinants.


4 February

Cramer's rule.  We finished Cramer's rule. This completes our discussion of determinants.

Physical vector spaces.  The word vector comes directly from Latin. It means "carrier." It was used in the seventeenth century in astronomy. The arrow from a planet to the Sun was thought of as "carrying" the planet around the Sun.

We looked at several common spaces of 3D physical vectors and spaces of coordinates: displacements, forces, and ℝ3. A displacement, or arrow, takes one from any initial point to a final point determined by the arrow. Forces, which are definitely not arrows, can still be described by them. Both displacements and forces also be described in terms of column vectors (or row vectors) in ℝ3. The three spaces are equivalent in the sense that vector addition and scalar multiplication behave exactly in the same way in the three spaces.


6 February

Vector spaces.  A vector space is a set V together with two operations, + and ·. V is required to be closed under these operations. This means that if u, v are in V, then u + v is in V; if c is a scalar, then c·v is in V. The operations have to satisfy the following rules.

Addition   Scalar multiplication
u + (v + w) = (u + v) + w   a·(b·u) = (ab)·u
Identity: u + 0 = 0 + u = u   (a + b)·u = a· u + b·u
Inverse: u + (-u) = (-u) + u = 0   a·(u + v) = a·u + a·v
u + v = v + u   1·u = u

Examples.  Displacements — addition (+), parallelogram law; scalar multiplication (·), stretching and/or reversing the direction of an arrow. Forces — addition (+), resultant; scalar multiplication (·), intensifying and/or reversing the direction of a force. ℝn — addition and scalar multiplication of column vectors. ℝm×n — addition and scalar multiplication of m×n matrices. C[a,b], the set of all scalar-valued continuous functions on [a,b] — addition (+), function addition, (f+g)(x) := f(x) + g(x); scalar multiplication (·), multiplying a function by a scalar, (c·f)(x) = c(f(x)). For more examples and further discussion, see chapter 2, section 1 of Heffron's online text.


9 February

Subspaces.  A nonempty subset U of V is a subspace of V if, under + and · from V, U is a vector space in its own right.

Checking to determine whether a particular set, together with operations +, ·, is a vector space can be tedious (cf. the list of axioms above). However, when we are checking to whether U is a subspace of a vector space V, things are simpler.

Theorem 2/9.1. A nonempty subset U of V is a subspace of V if and only if U is closed under + and · from V.
This theorem can be rephrased as a simple test that U must pass for it to be a subspace of V. Specifically, U will be a subspace of V if and only if these three conditions are satisfied:
  1. 0 is in U.
  2. U is closed under + .
  3. U is closed under · .
The first condition, namely that 0 is in U, does two things. First, it checks to see if U is nonempty. But, since 0 has to be in every vector space in any case, it serves to quickly eliminate sets without checking the closure properties.

Example 1. Determine whether U={(x1, x2, x3) ∈ ℝ3 | 2x1 + 3x2 - x3 = 0} is a subspace of R3.

Solution. Before we proceed to anything else, we need to be aware that for a vector x = (x1, x2, x3) to be in U the coordinates x1, x2, x3 have to satisfy the equation 2x1 + 3x2 - x3 = 0. So, for instance, (1,1,5) is in U, but (1,1,−5) is not in U. Let's go through our checklist.

  1. Is 0 = (0,0,0) in U? Yes — 2(0) + 3(0) - 5(0) = 0. The coordinates of 0 thus satisfy 2x1 + 3x2 - x3 = 0.
  2. Is U closed under + ? Suppose y = (y1, y2, y3) and z = (z1, z2, z3) are in U. Then y + z = (y1 + z1, y2 + z2, y3 + z3). Do the coordinates of y + z satisfy 2x1 + 3x2 - x3 = 0?

    2(y1 + z1) + 3(y2 + z2) − (y3 + z3) = 0 (?)

    Algebra:  (2y1 + 3y2 - y3) + (2z1 + 3z2 - z3) = (?)

    y, z in U: (0)+(0) = 0. ✔

  3. Is U closed under · ? Let c be a scalar and let x = (x1, x2, x3) be in U. Do the coordinates of c·x = (cx1,c x2, cx3) satisify our equation?

    2cx1 + 3cx2 - cx3 = 0 (?)

    Algebra:  c(2x1 + 3x2 - x3) = 0 (?)

    x is in U: c(0) = 0 ✔

Thus, U={(x1, x2, x3) | 2x1 + 3x2 - x3 = 0} is a subspace of ℝ3

Example 2. Determine whether W={(x1, x2, x3) ∈ ℝ3 | 2x1 + 3x2 - x3 = 1} is a subspace of R3.

Solution. No, because 0 = (0,0,0) is not in W.

Span.  We need to start with the definition of a linear combination of vectors. Let S = {v1 ... vk} be a set of vectors in a vector space V. A vector of the form c1v1 + ... + ck vk, where the cj's are scalars, is called a linear combination of the vj's. The span of S is the set of all linear combinations of vectors in S. That is,

span(S) := {c1v1 + ... + ck vk},

where the cjs are arbitrary scalars.
Proposition 2/9.2. The set span(S) is a subspace of V.
The polynomials of degree n or less, Pn = span{1, x, x2, ..., xn }. Since xk are continuous functions, the proposition above guarantees that they form a subspace of the continuous functions. Below is a list of vector spaces that are important, along with the operations of addition + and scalar multiplication ·

Important vector spaces

Displacements in space, forces, velocities, accelerations, etc.
n (real scalars) or ℂn (complex scalars) - n×1 real or complex matrices (i.e., columns; can also work with rows).
Pn = {a0 + a1 x + a2x2 + ... + anxn }, the polynomials of degree n or less.
Spaces of functions f : X -> scalars, where X can be anything. Think of f(x) as the ``x component of f.''
Various subspaces of the vector spaces of the type f : X -> scalars


11 February

Test for subspaces.  The test that we introduced for determining whether a subset U of a vector space V involved three steps: Is 0 in U? Is U closed under + ? Is U closed under · ? It's possible to combine the last two steps into a single step, where we check whether linear combinations of two vectors in are still in U. Specifically, we have the following "new" test:

  1. Is 0 is in U?
  2. If c, d are scalars and u, v are in U, then is c·u + d·v in U?
We applied this test to U = {f ∈ C[0,3] | f(0)=0 and f(3)=0}. The vectors here are actually functions. In particular, the "zero" of C[0,3] is the function z(x) ≡ 0 on the interval 0 ≤ x ≤ 3.

Span (continued).  We introduced some terminology: If a vector space V = span(S), we say that V is spanned by S, or V is the span of S, or S spans V. Thus, if S = {1, x, x2}, then the polynomials of degree 2 or less, P2, is spanned by S, and S spans P2.

When a vector space is spanned by a set, the information needed to describe the set is greatly reduced. The space P2 comprising all polynomials of degree n or less has an infinite number of polynomials in it. Describing as span{1, x, x2} reduces the information to knowing the coefficients of the three powers: 1, x, x2.

Another example is the set U={(x1, x2, x3) ∈ ℝ3 | 2x1 + 3x2 - x3 = 0}, which we have shown to be a subspace of ℝ3. Put the equation 2x1 + 3x2 - x3 = 0 in augmented form, [2 3 −1 | 0]. This matrix is row equivalent to [1 3/2 −1/2 | 0]. The leading (dependent) variable is x1, and the non-leading (independent0 variables are x2 and x3. Letting x2 = t1 and x3 = t2, the parametric form of the solution is

(x1 x2 x3)T = t1(−3/2 1 0)T + t2(1/2 0 1)T

Thus the set of solutions is the span of S = {(−3/2 1 0)T, (1/2 0 1)T}.

13 February

Subspaces associated with a matrix.  Let A be in ℝm×n (or ℂm×n). There are four important subspaces associated with a matrix.

  1. The null space of A. null(A) = {x ∈ ℝn   |   Ax = 0}. This is a subspace of ℝn.
  2. The null space of AT. null(AT) = {y ∈ ℝm   |   ATy = 0}. This is a subspace of ℝm.
  3. The column space of A. ColSpace(A) = span{columns of A} = {y ∈ ℝm   |   Ax = y}. This is a subspace of ℝm.
  4. The row space of A. RowSpace(A) = span{rows of A}. This is a subspace of ℝ1×n.

Linear independence and dependence.  We say that a set of vectors
S = {v1, v2, ... , vk}
is linearly independent (LI) if the equation
c1v1 + c2v2 + ... + ckvk = 0
has only c1 = c2 = ... = ck = 0 as a solution. If it has solutions different from this one — i.e., non trivial (non zero) — then the set S is said to be linearly dependent (LD).

Examples. 

  1. The columns of the matrix A =
    1 1 2  0
    2 4 2  4
    2 1 5 -2
    
    are linearly dependent because 3col1col2col3 = 0.

  2. Determine whether the set {1 - x, x - x2, 2x + x2} is LI or LD.

    Solution First form the equation
    (∗) c1(1 -x) + c2(x - x2) + c3(2x + x2) ≡ 0, the polynomial that's identically 0 (i.e., 0 for all x).
    Since the polynomial is 0 for all x, we may use x = 0 in the equation. This gives us c1(1 - 0) + c2(0 - 02) + c3(2· 0 + 02) = 0. Thus, we see that c1 = 0. Using this in (∗) gives us
    c2(x - x2) + c3(2x + x2) ≡ 0.
    For x ≠ 0, divide this equation by x to get c2(1 - x) + c3(2 + x) ≡ 0. By continuity this actually holds for all x, including x = 0. Choosing x = −2 in this equation gives 3c2 + 0c3 = 0, so c2 = 0. Finally, take x = 1 in the equation left over to get c3 = 0. It follows that the set is LI.

  3. Other sets.
    1. {1, x, x2, ..., xn } is linearly independent.
    2. {1 + x, 1 - x, 1} is linearly dependent.
    3. { i, j, k} is linearly independent.
    4. { (1,0,2), (0,1,3) } is linearly independent and { (1,-1,-1), (1,2,7), ((2,-1,1) } is linearly dependent.
    5. {ex, e2x, e3x} is linearly independent.

Inheritance properties of LI and LD sets. . Every subset of a linearly independent set is linearly independent. Every set that contains a linearly dependent set is linearly dependent.


16 February

Basis for a vector space.  We have discussed the concepts of span and of linear independence and dependence. We now want to combine them. Suppose that V is a vector space and B = {v1, v2, ... , vn} is a set of vectors in V. Then, the following is true.

Theorem 2/16.1.  If B spans V and is linearly independent, then every vector v in V can be written in one and only one way as a linear combination of vectors in B. That is,
(∗)  v = c1v1 +...+ cnvn
where the coefficients are unique.
Proof. Since B spans V, we have scalars c1, ..., cn such that the representation above holds. We want to show that this representation is unique - i,e,, no other set of scalars can be used to represent v. Keeping this in mind, suppose that we have a second representation
(¶)  v = d1v1 +...+ dnvn,
where the coefficients are allowed to be different from the ck's. Subtracting the two representations for v yields
0 = (c1 − d1) v1 +...+ (cn − dn) vn.
Now, B is LI, and so the last equation implies that c1 − d1 = 0, c2 − d2 = 0, ..., cn − dn = 0. That is, the c's and d's are the same, and so the representation is unique. ◼
The coefficients in (∗) completely and uniquely describe the vector v. They are coordinates for V. The spanning property of B allows us to represent a vector and linear independence makes that representation unique. A set B that both spans V and is linearly independent will be called a basis for V. We will discuss coordinates in greater detail later on.

Dimension of a vector space.  Vector spaces fall into two categories, depending on the largest size of a linearly independent set. If a vector space V has no limit to the size of its linearly independent sets, it is said to be infinite dimensional. If the largest number of vectors in any linearly independent set is a finite number n, then V is said to be finite dimensional and the dimension of V is n. 2D and 3D displacements are finite dimensional spaces, as is P3 = span{1, x, x2, x3}. The 2D displacements have dimension n = 2, the 3D, n = 3, and the polynomial space, n = 4. The space C[−π,π] is infinite dimensional, because it contains {1, cos(x), sin(x), cos(2x), sin(2x), ..., cos(Nx), sin(Nx)}, which is linearly for all N, no matter how large.

When V is finite dimensional, its LI sets cannot be arbitrarily large. Suppose that n is the dimension of V; that is, n is the maximum number of vectors an LI set in V can have. A linearly independent set S in V having n vectors in it will be called maximal.

Theorem 2/16.2.  Let V be a finite dimensional vector space. Then any maximal linearly independent set of vectors
S = {v1, v2, ... , vn}
is a basis for V. Conversely, any basis is a maximal linearly independent set.
Proof. We start with showing the if S is maximal, then it is a basis. Add any vector v in V to the maximal set S to form the new set
{v, v1, v2, ... , vn}
This augmented set is LD, because it has n+1 > n vectors in it. Thus we can find coefficients c, c1, c2, ... , cn, at least one of which is not 0, such that this holds:
(∗) cv + c1v1 + ... + cn vn = 0.
There are two possibilities. Either c is 0 or c is not 0. In case c = 0, we have c1v1 + ... + cnvn = 0. The set S is LI, so this implies c1 = c2 = ... = cn = 0. But this means all of the coefficients vanish, contradicting the fact that one of them does not vanish. The only possibility left is that c is not 0. We can then divide by it in our previous equation and rewrite that equation as
v = (- c-1c1)v1 + ... + (- c-1cn)vn.
This shows that S spans V. Since S is LI, it is a basis.

(We include the proof below for completeness.) We now assume that S is a basis, but possibly not maximal. We want to show that S is maximal. Suppose that
T = {w1, w2, ... , wm}
is a maximal set, with m > n. Since S is a basis, we may represent every vector in T as a linear combination of the vectors in S. Thus, we have scalars aj,k such that
wk = a1,kv1 +...+ an,kvn
Form the following linear combination of vectors and set the result equal to 0.
(∗∗) x1w1 + x2w2 + ... + xmwm = 0
Because the w's are linearly independent, the only solution to (∗∗) is x = 0. On the other hand, we can use the preceeding equation to replace the w's with v's. With a little work, the result can be expressed in matrix notation as follows:
(Ax)1v1 +...+ (Ax)nvn = 0.
Since the v's form a basis, they are linearly independent, and so (Ax)j = 0, for j = 1, ..., n. Putting this completely in matrix form we have Ax = 0. However, A is an n×m matrix, with m > n — i,e., the "number of equations n is smaller than the number of variables m." Thus, Ax = 0 has solutions x0. This is a contradiction, because the linear independence of the v's implies that the only solution is x = 0. Thus, m = n and S is maximal. Our proof is complete. ◼

This has an important consequence that needs to be emphasized.
Corollary 2/16.3.  Let V be finite dimensional. The dimension of V is the number of vectors common to a basis.
For example, Pn has {1, x, ..., xn} as a basis, so dim(Pn) = n+1. Also, the dimension of ℝn is n. The definition of dimension is frequently defined as the number of vectors in a basis, rather than the maximal size of a set of linearly independent vectors. The two definitions are equivalent.

We also discussed methods for finding bases for the subspaces associated with a matrix — the null space, column space, and row space; these may be found in my notes on Methods for Finding Bases.


18 February

Bases and coordinates.  Assigning coordinates amounts to providing a correspondence
v ↔ (c1, c2, ... , cn)
that is 1:1, onto, and preserves vector addition and scalar multiplication. This is exacly what a basis does. Let B = {v1, v2, ... , vn} be a basis for a vector space V; i.e., B is both LI and spans V. We also assume that the basis is ordered, in the sense that we keep track of which vector is first, or second, etc. Theorem 2/16.1 gives us a way to assign coordinates to V. The correspondence,
v= c1v1 +...+ cnvn ↔ (c1 c2 ... cn)T,
that it sets up is both 1:1 and onto — every  v corresponds to one column and every column corresponds to one  v. It is also easy to show that this correspondence preserves addition and scalar multiplication, which are the properties needed in defining "good" coordinates for a vector space. We say that the column vector
[v]B = (c1 c2 ... cn)T
is the coordinate vector for v, and that c1, c2, ..., cn are the coordinates for v. A correspondence between two vector spaces U and V,
uv,
that is 1:1, onto, and preserves vector addition and scalar multiplication is called an isomorphism between U and V, and the two spaces are said to be isomorphic.

The word isomorphism comes from two Greek words, "isos," which means "same", and "morphy," which means "form." As far as vector space operations go, two isomorphic vector spaces have the "same form" and behave the same way. Essentially, the spaces are the same thing, just with different labels. For example,a basis in one space corresponds to a basis in the other. Indeed, any property in one space that only involves vector addition and scalar multiplication will hold in the other. This makes the following theorem, which is a consequence of what we said above concerning coordinates, very important.

Theorem 2/18.1.  Every n-dimensional vector space is isomorphic to ℝn or ℂn, depending on whether the scalars are real or complex.
Examples. The space P2 of polynomials of degree 2 or less has B = { 1, x, x2 } as a basis, and so it is three dimensional (B has three vectors). We take the scalars to be real. Relative to this basis, we have
[p]B = [ a1 + a2x + a3x2 ]B = [a1 a2 a3]T.
Thus, for example, we have these:
[1-x+x2]B = [1 -1 1]T
[-3x+4-2x2]B = [4-3x-2x2]B = [4 -3 -2]T
[(2-x)2]B = [4-4x+x2]B = [4 -4 1]T
Changing the ordering of the basis vectors changes the ordering of the coordinates. Using the basis C = { x2 1, x }, we have
[-3x+4-2x2]C = [-2x2+4-3x]C = [-2 4 -3]T,
which is a reordering of the coordinate vector relative to B. Be aware that order matters!

More examples and a further explanation of the notation may be found in my notes on Coordinate Vectors and Examples.


20 February

More examples of bases.  Let U be the subset of Pn comprising all polynomials such that p(1)+p'(1)=0 and p(2)=0. Recall that in assignment 4 you showed that U is a subspace of P3. To find a basis for U, first assume that p(x) = a0 + a1x+ a2x2 + a3x3 is in U. We apply the two conditions on p to get two equations

a0 + 2a1 + 3a2 + 4a3 = 0 and a0 + 2a1 + 4a2 + 8a3 = 0.

Solving these equations leads to the {2 − x, 8 − 4x2 + x3} being a basis for U.

The second example was involved ``tent'' functions and linear splines. We'll skip the details here.

Change of basis.  Frequently we want to use a different basis for a vector space instead of the one we started with. This will be especially important when we diagonalize matrices. Let B = {v1 ... vn} and D = {w1 ... wn} be ordered bases for a vector space V. Suppose that we have these formulas for v's in terms of w's:

vj=A1,jw1 + A2,jw2 + ... + An,jwn = Σi Ai,jwi

(Note that the sum is over the row index for the matrix A.) In practice, this means that we know the basis vectors in B, the v's, in terms of the D basis vectors, the w's.

[vj]D = (A1,j A2,j ... An,j)T = colj(A),  j=1,...,n.

We can represent any vector u in terms of the basis B (the v's) and D (the w's),
u = b1v1 +...+ bnvn and u = d1w1 +...+ dnwn.
The coordinate vectors relative to B and D are thus
[u]B = [b1,..., bn]T and [u]D = [d1,..., dn]T.
Now, using the representation of u in terms of B, we can take D coordinates to get
[u]D = b1[v1]D +...+ bn[vn]D = b1 col1(A) +...+ bn coln(A).
Applying the ``basic matrix trick'' to the right side above, we arrive at the change-of-basis formula

[u]D = A[v]B.


23 February

Change of basis examples.  We will work with the following two bases for P2:
B = { 1, x, x2 } and D = { 1 + x, 1 − x2, 1 + x + x2 }.
We will let p1(x) = 1 + x, p2(x) = 1 − x2, and p3(x) = 1 + x + x2. Suppose that we have that a polynomial
p = −5p1 + 3p2 + 4p3 — hence [p]D = (−5 3 4)T.
To find the B coordinates, note that [p]B = −5[p1]B + 3[p2]B + 4[p3]B = A[p]D = A(−5 3 4)T, where A =

1  1 1
1  0 1
0 −1 1
Thus, [p]B = (2 −1 1)T, and so p(x) = 2 − x + x2.

As a second example, we take the vector space to be P1. We will work with three bases this time. Suppose B = { 1 + x, 1 −1 x} and D = {2 − x, 1 + 3x}. The third basis — the one B and D are given in — is E = { 1, x}. Suppose that
p(x) = 4(2 − x) − (1 + 3x), so [p]D = (4 −1)T.
We want the B coordinates of p, [p]B = (b1 b2)T. Now, in the B and D coordinates we have
p(x) = b1(1 + x) + b2(1 − x) = 4(2 − x) − 3(1 + 3x).
We now find the E coordinates of all of both sides.
[p]E = b1(1 1)T + b2(1 −1)T = 4(2 − 1)T − 3(1 3)T = (5 − 13)T. To find the b's, we need to solve A[p]B = (5 − 13)T, where A =

1  1
1 −1
Doing so gives us [p]B = (−4 9)T.

Functions.  A function f: D → R is a rule that assigns to each x ∈ D, the domain of f, a y ∈ R, the range. The function f is one-to-one if and only if f(x1) = f(x2) has x1 = x2 as its only solution. For example, if f:ℝ → ℝ, f(x) = x3, then f is one-to-one. On the other hand, f(x) = x2 is not one-to-one, because x12 = x22 does not imply x1 = x2, since except when x12 = x22 = 0. A function f is onto if and only if every y ∈ R has the form y = f(x) for some x ∈ D.

The term range is used in two ways by different mathematicians. The first is the set in which f takes on its values — f(x) = x2 takes on values in ℝ, just not all of ℝ. The second is all the values the function actually hits. For f(x) = x2, this is the set [0,∞). We will have more to say about this next time.


25 February

Linear transformations.  A mapping L:V → W, where V and W are vector spaces, is said to be a linear transformation if it satisfies these properties.

  1. Homogeneity L[au] = aL[u]
  2. Additivity L[u + v] = L[u] + L[v]

    Equivalently, we can combine these two properties into a single property:

  3. Linearity L[au + bv] = aL[u] + bL[v]
These properties mimic familiar ones for derivatives and integrals in calculus. The first property is the "constant multiplier rule" and the second is the "sum rule."

Simple properties.

Examples of linear transformations.  There are many examples of linear transformations. Here are a few of them.
  1. The derivative — V = C(1)(ℝ), W = C(ℝ),  L[f] = df/dx
  2. Linear differential operators — V = C(2)(ℝ), W = C(ℝ),  L[u] = x2u′′ + (2x+1)u′ + 3u
  3. Rotations — V = W = 2D-displacements,  Rθ[v] = vθ, which is v rotated through an angle θ
  4. Matrix multiplication — V = ℝn, W = ℝm,  L[v] = Av, where A is an m×n matrix.
Matrix representations. 

Theorem 2/25.1 (Matrix associated with L) Let V and W be finite dimensional, and let B = {v1, ... , vn} and D = {w1, ... , wm} be bases for V and W, respectively. If L:V → W is a linear transformation, then there is a unique m×n matrix A such that w = L[v] holds if and only if A[v]B = [w]D.
Proof Let v = c1v1 + c2v2 + ... + cnvn, so [v]B = [c1, ... , cn]T. From the first of the simple properties above, we see that
[ L[v] ]D = c1[ L[v1] ]D + c2[ L[v2] ]D + ... + cn[ L[vn] ]D
Now, by our "basic matrix trick," we can write this as a matrix product,
[ L[v] ]D = A[v]B, where
A = [ [ L[v1] ]D, ... , [ L[vn] ]D ]
is the m×n matrix we wanted. This matrix is unique because its columns are the unique coordinates for L[vj] ] relative to the basis D.
Consider the following problem. Let V = W = P2 have the common basis B = D = {1,x,x2}. Suppose that L:P2P2 is a linear transformation given by

L[p] = x2p'' + (2x+1)p' + 3p,

Find the matrix A that represenets L. In addition, find the solution to L[p] = 18x2 - x + 2.

To find A, we first find the output of L applied to each basis vector; that is, L[1], L[x], and L[x2]. Doing this, we obtain L[1] = 3, L[x] = 1 + 5x, and L[x2] = 2x + 9x2. By the construction in the theorem, the kth column of A is the coordinate vector [ L[vk] ]D. Consequently, we have
[ L[1] ]D = [3 0 0]T
[ L[x] ]D = [1 5 0]T
[ L[x2] ]D = [0 2 9]T
We have now found that the matrix A =

3 1 0
0 5 2
0 0 9
To solve L[p] = 18x2 - x + 2, we go over to the matrix form of the equation,
A[p]B = [18x2 - x + 2]B = [2 -5 18]T.
Solving this equation, we get
[p]B = A-1[2 -1 18]T = [1 -1 2]T
Putting this back in terms of polynomials, we arrive at p(x) = 2 - x + x2

27 February

Subspaces associated with a linear transformation L : V → W

Examples

  1. Matrices  Consider the 2×4 matrix M =
    1 -1 2 3
    1  1 1 4
    
    We let L be the transformation L[v] = Mv, where v is in R4. For this problem, V = R4. The co-domain or set of outputs W is R2. Here, we also have image(L) = R2. The null space is the set of all v for which Mv = 0. Using row-reduction, one can show that
    null(L) = span{[-3 1 2 0]T, [-7 -1 0 2]T}

  2. ODEs  Consider a differential operator of the form L[p] = (1−x2′′−2xp′ = 0. In this case, we take the the domain V and the co-domain to be to be P2. We want to find bases for the null space, null(L), and the image space, image(L). To do this, apply the representation theorem to get the matrix of L relative to a convenient basis for both V = W = P2; the easiest to use is B = {1,x,x2}. Let's do this:
    L[1] = 0, L[x] = −2x, and L[x2] = 2 − 6x2
    [L[1]]B = (0 0 0)T, [L[x]]B = (0 −2 0)T, and [L[x2]]B = (2 0 −6)T

    Hence, the matrix for L is AL =

    0  0  2
    0 −2  0
    0  0 −6
    

    Proceeding as in the previous example, we see that the basis for
    NullSp(AL) = {(1 0 0)T)} and ColSp(AL) = {(0 −2 0)T,(2 0 −6)T}.
    The last step is to put everything in terms of polynomials using the basis B = {1,x,x2} and the bases of coordinates that we just found. The final answer is this:
    The basis for null(L) is {1}, and the basis for image(L) is {−2x, 2 − 6x2}.

Combinations of linear transformations

Sums K+L is defined by (K+L)[v] = K[v] + L[v].

Scalar multiples cL is defined by (cL)[v] = c(L[v]).

Products If K : V → U and L : U → W are linear, then we define LK via LK[v] = L[K[v]]. (This is composition of functions). The transformation defined this way, LK, is linear, and maps V → W. Note: LK is not in general equal to KL, which may not even be defined.

Inverses Let L : V → V be linear. As a function, if L is both one-to-one and onto, then it has an inverse K V → V. One can show that K is linear, and LK = KL = I, the identity transformation. We write K = L-1.

Associated matrices Recall that if B = {v1, ... , vn} and D = {w1, ... , wm} be bases for V and W, respectively, then the matrix associated with the linear transformation L : V →; W is
ML = [ [ L[v1] ]D, ... , [ L[vn] ]D ]
Since each of the combinations listed above is still a linear transformation, it will have a matrix associated with it. Here is how the various matrices are related.

MK + L = MK + ML

McL = c ML

MLK = ML MK

ML-1 = (ML)-1

6 March

Change-of-basis and linear transformations.  Let L : V → V be linear. Here L maps V into itself. We want to look at what happens when to the matrix of L when we make a change of basis in V. Let V have bases
B = {v1, ... , vn} and D = {w1, ... , wn},
and let S = [D-basis]B-coords be the matrix that takes D-coordinates into B-coordinates. Of course, S−1 does the reverse — it takes B-coordinates to D. Suppose that we know the matrix of L relative to B, AB, and that we want to find the AD, the matrix of L in D-coordinates. Theorem 2/25.1 implies that the equation Lx = y is completely equivalent to both of these equations:
AB[x]B = [y]B and AD[x]D = [y]D.
Starting with what we know, the left equation above, we change to D-coordinates on both sides of that equation. This yields
ABS[x]D = S[y]D.
Inverting S on the right-hand side above then gives us S−1ABS[x]D = [y]D = AD[x]D, which happens if and only if
AD = S−1ABS.

Similar matrices.  The two matrices AB and AD both represent L, but in different bases. We say that an n×n matrix B is similar to an n×n matrix B if there exists an invertible matrix S such that B = S-1AS. Simple properties of similarity are listed below. We remark that these properties together make similarity an equivalence relation.

  1. B is similar to A if and only if A is similar to B. (Thus, we can simply say that A and B are similar matrices.)
  2. A is similar to A.
  3. If A is similar to B, and B is similar to C, then A is similar to C.

Trajectories of a velocity field.  Suppose that we are given a time-independent velocity field for a 2D fluid, V(x). Assume that V is a linear function of x, so that V(x) = L[x], where L is a linear transformation taking 2D to 2D. We want to find the trajectories of the fluid particles, dx/dt = V(x) = L[x], given that we know the initial position x(0). Let B = {i,j} be the usual basis for 2D. Relative to this basis L will have a matrix AB. so our problem becomes [dx/dt]B = A[x]B. Since the basis B is time-independent, [dx/dt]B = d[x]B/dt. If we let X = X(t) = [x]B=[x1 x2]T, we arrive at the system
dX/dt = ABX, where X(0)= [x(0)]B.
To make the situation more concrete, we will assume that AB =

1  4
1 -2
When this system is written out, it looks like
dx1/dt = x1 + 4x2
dx2/dt = x1 - 2x2
The idea here is to switch to a basis where the system decouples. To accomplish this, we will use this basis D = {i+4j, -i+j}, which is {[4 1]T, [-1 1]T} in B-coordinates. The matrix S = [D-basis]B-coords is given by S =
4 -1
1  1
The matrix of L relative to D is AD = S−1ABS =
2  0 
0 -3 
Letting Z = [x]B' = [z1 z2]T, we have the new system dZ/dt = A'Z, with Z(0) = [x(0)]B'. In the new coordinates the system decouples and becomes
dz1/dt = 2z1
dz2/dt = - 3z2
This decoupled system is easy to solve. The result is
z1(t) = e2tz1(0)
z2(t) = e-3tz2(0)
Transforming back to the original coordinates, we have that X = SΛ(t)S-1X(0), where Λ(t) =
exp(2t)   0  
0         exp(-3t)
This explicitly solves the problem. However, we still have to explain how to find coordinates that decouple the system. Doing this is closely connected with solving what are called eigenvalue problems.

The Eigenvalue problem.  We say that a scalar λ is an eigenvalue of L : V → V if there is exists a vector v in V, with v0, such that L[v] = λ v. The vector v is called an eigenvector of λ. We let Eλ be set eigenvectors associated with λ, along with 0. Eλ is a subspace of V that is called the eigenspace of λ. The eigenvalue problem is to find all eigenvalues and corresponding eigenvectors for L.

In the example we just did, the matrix AB =

1  4
1 -2
had two eigenvalues: λ = 2, whose corresponding eigenvectors are multiples of [4 1]T; and λ = −3, with multiples of [-1 1]T being the corresponding eigenvectors. Using the basis D = {[4 1]T, [-1 1]T}, which comprises only eigenvectors of AB, the "decoupled" matrix representing L is the diagonal matrix AD =
2    0  
0   −3

Diagonlizable linear transformations.  A linear transformation L : V → V is diagonalizable if and only if there is a basis D for V comprising only eigenvectors of L. When this happens AD, the matrix of L relative to the basis D will be a diagonal matrix. Conversely, if there is a basis relative to which the matrix of L, AD, is diagonal, then that basis will be composed of eigenvectors. Being able to "decouple" a linear system related to a linear transformation L is thus equivalent to being able to diagonalize L.


9 March

Examples of eigenvalue problems and matrix diagonalization.  See my notes on Diagonalization.



Vector Analysis

11 March

Piecewise smooth curves.  The motivation for defining scalar and vector line integrals is given in sections 14.1 and 14.2 of the Cain-Herrod text. We discussed parametrizing various curves — circles, lines, spirals, and others. In addition, we defined smooth curves (continuously differentiable and having nonzero velocity) as well as piecewise smooth curves, which are several smooth curves joined together at endpoints. Finally, we discussed the sense of traversal of a piecewise smooth curve and finished with changing the parametrization for smooth curves.


13 March

Scalar and vector line integrals.  Scalar line integrals are used to represent scalar quantities, such as the mass or charge of a wire when the appropriate density is known.

Both types of integrals are discussed in sections 14.1 and 14.2 of Cain-Herrod. There is, however, an unfortunate notational choice in the text. The differential dr is used to denote both the scalar arclength differential and the vector displacement differential. This is bad notation because it does not distinguish between a scalar (arclength) and a vector (displacement); normally, dr (or dx) is used to denote the vector displacement and ds is used for the arclength.

The arclength parameter.  For a smooth curve parametrized by r = r(t), where a ≤ t ≤ b, the velocity is v(t) = dr/dt. Of course, since the curve is smooth, v(t) ≠ 0, and therefore the speed |v(t)| ≠ 0. The arclength of the section of the curve starting at r(a) and ending at r(t) is
s(t) = ∫at |v(τ)|dτ.
Now, ds/dt = |v(t)| > 0. We may thus solve s = s(t) for t = t(s). The arclength s then becomes a parameter, with ds being its differential — a quantity with geometrical meaning. An integral of scalar quantity f(r) over a curve is thus represented as
Cf(r)ds.
When a parametrization r r(t) is given, the integral is computed via the formula
Cf(r)ds = ∫ab |v(t)|dt.
See Cain-Herrod and your class notes for examples.


23 March

Physical interpretations vector line integrals.  If F(r) is a vector force field, e.g. gravity, given in a region in 2D or 3D space, then the work done by the force F in displacing an object dr is F·dr. W, the total work done in moving the object over a piecewise smooth curve C, is then the line integral
W = ∫ C F·dr. If instead of a force F we are dealing with a fluid's velocity field V(r) and the curve C is closed, the line integral
C V·dr
is called the circulation over C. Physically, the circulation is a measure of the tendency of a fluid to rotate. We did a number of line integrals as examples.

Path independence.  A line integral ∫ C F·dr, where C joins points A and B in a region R, is said to be path independent in that region if the value of the integral is the same for all paths from A to B lying in R. A necessary and sufficient condition for path independence is that the vector field F be of the form
F = ∇f := ∂f/∂x i + ∂f/∂y j + ∂f/∂z k
for some scalar function f. The operator ∇ is called the gradient.


25 March

Surfaces.  We discussed ways of representing surfaces, primarily looking at parametric representations. We covered §I in my notes on Surfaces.

27 March

Surfaces area elements and surface integrals.  We covered the material at the start of §II in my notes on Surfaces, deriving the formula for the element dS (or dσ) of surface area.

30 March

Density (scalar) integrals.  We discussed computing surface area elements, standard and unit normals to surfaces, and related topics. These can be used to compute two types of integrals. The first is an integral that represents area, total mass or charge, or some other geometrical or physical quantity. These have the form ∫∫Sf(r)dS, which are computed by using a parametric representation for the surface to convert the surface integral to a standard double integral.

Flux (vector) integrals.  These integrals require motivation. Consider the steady state velocity field V(x) of a fluid. We want to calculate the amount of fluid crossing a surface parametrized by r = r(u,v). Let ru and rv be partials of r with respect to the parameters u and v. We consider an element of surface area, shown below as the base of the parallelepiped. Our first step is to calculate the fluid crossing this surface element. In time t to t+dt, the volume of fluid crossing the base of the parallelepiped equals its volume,
(Vdt)·ru×rv dudv.

      Vdt
              rudu                         rvdv

The mass of the fluid crossing the base in time t to dt is then density×volume, or
Vdt)·ru×rv dudv.
Thus the mass per unit time crossing the base is F·N dudv, where F = ρV, and N = ru×rv is the standard normal. Recall that the area of the surface element is dS = |N|dudv. Consequently the mass per unit time crossing the base is F·n dS, where n is the unit normal. Integrating over the whole surface then yields

This surface integral is called the the flux of the vector field F. Such integrals are also referred to as vector surface integrals.


1 April

Green's Theorem.  Green's Theorem relates a line integral over a simple, closed curve to a double integral over the region enclosed by the curve — a 1D integral to a 2D integral. To state Green's theorem, we need to discuss simple, closed curves. These are closed curves, like circles, but the do not intersect themselves. Rectangles, triangles, circles, and ellipses are simple closed curves; figure eights are not. Simple closed curves divide the plane into two nonoverlapping regions, one interior and the other exterior. It forms the boundary of both regions. We will consider simple closed curves that are piecewise smooth, which just means that we are allowing a finite number of corners. We also say that a simple closed curve is positively oriented if it is travered in the counterclockwise direction. Here is the statement of Green's Theorem:

Green's Theorem Let C be a piecewise smooth simple closed curve that is the boundary of its interior region R. If F(x,y) = p(x,y)i + q(x,y)j is a vector-valued function that is continuously differentiable on and in C, then

Further discussion and examples are given in sections 17.2 and 17.3 of Cain/Herrod. We did examples verifying Green's theorem and computing areas; in particular, we showed how to find the area of an ellipse. Section 17.3 of Cain/Herrod shows how to find the area of a polygon using Green's theorem.


3 April

Green's Theorem and path independence.  If a vector field F(x,y) = p(x,y)i + q(x,y)j satisfies ∂q/∂x = ∂p/∂y in a region enclosed by a simple closed curve C, then by Green's Theorem ∳C F·dr = 0 — provided p, q, and their first partials are continuous in the region enclosed by C. When the continuity requirements are violated, this may no longer be true. In class, we looked at the vector field

F = (−y)/(x2+y2)i + x/(x2+y2)j, where p = (−y)/(x2+y2) and q = x/(x2+y2).

Taking the derivatives of p and q, we see that ∂q/∂x = ∂p/∂y. Let C be a circle of radius r = a. If we directly calculate the line integral involved we obtain

C F·dr = 2π

rather than 0. Green's Theorem doesn't apply because p and q are discontinuous at the point (0,0), which is inside the circle. In class, for any simple closed curve C enclosing a region R containing (0,0), we showed how to break up R in a way that allows us to still use Green's Theorem to find the line integral over C. In fact, we showed that for such a curve, the line integral has the value 2π. Examining this case leads to Green's Theorem for multiply connected regions, which we touched on briefly.

6 April

The curl and Stokes's Theorem.  We will assume that F has continuous partial derivatives. The curl is then defined by


There is a useful physical interpretation for the curl. Suppose that a fluid is rotating about a fixed axis with angular velocity ω. Define ω to be the vector with magnitude ω and with direction along the axis of rotation. The velocity of an element of the fluid located at the position with radius vector x is v(x) = ω×x. With a little work, one can show that ω = ½∇×v. Thus one half of the curl of the velocity vector v is the vector ω mentioned above.

Stokes's Theorem relates the line integral over a simple closed curve that forms the boundary of a (one-sided) surface. As with Green's Theorem, Stokes's Theorem relates a line integral, which is a one dimensional integral, to a certain surface integral. In fact, Green's Theorem is actually a special case of Stokes's Theorem, which we now state.

Stokes' Theorem  Let S be an orientable surface bounded by a simple closed positively oriented curve C. If F is a continuously differentiable vector-valued function defined in a region containing S, then

Example  Verify Stokes's Theorem for the vector field F(x) = 2yi + 3xj - z2k over the surface S, where S is the upper half of a sphere x2+y2+z2 = 9 and C is its boundary in the xy-plane, the circle x2+y2 = 9. C is traversed counterclockwise.

We will first compute the line integral over C. In the xy-plane, C is parameterized via
r(t) = 3 cos(t) i + 3 sin(t) j,   0 ≤ t ≤ 2π,
and so we have:

dr = (- 3 sin(t) i + 3 cos(t) j)dt
F(r(t)) = 2·3 sin(t)i + 32 cos(t)j - 02k = 2·3 sin(t)i + 32 cos(t)j
F·dr = (- 18 sin2(t) + 27 cos2(t))dt
CF·dr = ∫0(- 18 sin2(t) + 27 cos2(t))dt = 9π
We now turn to finding the surface integral. ∫∫S ∇×F·n dσ. The normal compatible with the orientation of C is n = r/|r| = r/3. Thus, on the surface of the hemisphere S, we have
n = x(θ,φ)/3 = (3 sin(θ)cos(φ) i + 3 sin(θ)sin(φ) j + 3 cos(θ) k)/3,
and hence,
n = sin(θ)cos(φ) i + sin(θ)cos(φ) j + cos(θ) k9
where 0 ≤ θ ≤ ½π and 0 ≤ φ ≤ 2π. Also, the area element is dS = 32sin(θ)dφdθ. Moreover, it is easy to show that &nabla×F = k. We are now ready to do the surface integral involved:
∫∫S ∇×F·n dS = ∫∫S k·n dS
∫∫S ∇×F·n dS = ∫00½π cos(θ) 32 sin(θ)dφdθ
∫∫S ∇×F·n dS = 9π
Since both terms in Stokes's Theorem have the same value, we have verified the theorem in this case.

6 April

Gauss's (Divergence) Theorem.  Gauss's (or the Divergence) Theorem relates a flux integral for a vector field F over a closed surface to an integral over the volume enclosed by the surface. Let F(r)=p(x,y,z)i + q(x,y,z)j +r(x,y,z)k be a vector field with continuously differentiable components. We define the divergence of F to be the scalar-valued function

For example, if F = (x2y -zx)i + (xz -5y)j +z3y-4xk, then ∇·F = 2xy − 5 + 3z2y. Like the curl, the divergence of F has a physical interpretation in terms of fluids. This will be made clearer later. Here is the statement of the Divergence Theorem.

Gauss's (Divergence) Theorem  Let V be region in 3D bounded by a closed, piecewise smooth, orientable surface S; let the outward-drawn normal be n. Then,
Example  Verify the Divergence Theorem for the surface integral ∫∫S F·ndσ, where F = 3xi+yj+2zk and S is the surface of the closed cylinder (including caps) x2+y2 = 16, 0 ≤ z ≤ 5. The normal is outward drawn.

To do this we must compute both integrals in the Divergence Theorem. We will first do the volume integral. It is easy to check that ∇·F = 3+1+2=6. Hence, we have that
∫∫∫V·FdV = ∫∫∫V6dV = 6π·42·5 = 480π

The surface integral must be broken into three parts: one for the top cap, a second for the curved sides, and a third for the bottom cap.
∫∫S F·ndS = ∫∫top F·ndS + ∫∫sides F·ndS + ∫∫bottom F·ndS
The outward normals for the top and bottom caps are k and −k, respectively. For the top (z = 5), we are integrating F(x,y,5)·k = 2·5 = 10, and for the bottom (z = 0), F(x,y,0)·(−k) = −2·0 = 0. Hence, we have
∫∫top F·ndS = ∫∫topdS = 10π42 = 160π
∫∫bottom F·ndS = ∫∫bottom 0dS = 0
The integral over the curved sides will require a little more effort. The outward normal (see my notes, Surfaces, pg. 5) and area element are, respectively,
n = cos(θ)i + sin(θ)j and dS = 4dθdz.
In addition, on the curved sides
F(4cos(θ),4sin(θ),z) = 12cos(θ)i + sin(θ)j + 2zk, so F·n = 12cos2(θ) + 4sin2(θ).
The surface integral over the curved sides is then given by
∫∫sides F·ndS = ∫050 (12cos2(θ) + 4sin2(θ))4dθdz = 5·4(12π+4π)= 320π.
Combining these three integrals, we obtain
∫∫S F·ndS = 160π+320π + 0 = 480π,
which agrees with the result from the volume integral. Thus we have verified the Divergence Theorem in this case.

Equation of continuity for fluids.  Suppose that in a region a fluid has a velocity field v(x,t) and density ρ(x,t), and that there are no sources or sinks in the region. Earlier (30 March) we showed that the amount of fluid crossing a surface in the direction n per unit time is the flux,
Φ = ∫∫Sρv·ndS.
If S is a closed surface forming the boundary of a volume V, then Φ is the negative of the total rate of change of mass in the fluid in V. Thus, we have the equation
∫∫Sρv·ndS = − d/dt ∫∫∫VρdV = − ∫∫∫V∂ρ/∂t dV.
If we use the divergence theorem to replace the surface integral by a volume integral, we obtain
∫∫∫V·v)dV = − ∫∫∫V∂ρ/∂t dV,
and, consequently, that
∫∫∫V (∇·v)+∂ρ/∂t) dV = 0
holds for every choice of V within the region under consideration. If we take V to be a small sphere of radius ε and center x, then the limit as ε tends to 0 of
V-1∫∫∫V (∇·v)+∂ρ/∂t) dV
is ∇·v)+∂ρ/∂t. On the other hand, this limit is of course 0. Therefore,
·v)+∂ρ/∂t = 0.
This partial differential equation is called the equation of continuity.


Updated 4/14/2009.