## Practice Exam 2 M314

[1] (6 points) Let A be an n x n matrix. If the equation Ax = 0 has only the trivial solution, do the columns of A span Rn? Why or why not?

Answer: To say that the columns of A span Rn is the same as saying that Ax = b has a solution for every b in Rn. But if Ax = 0 has only the trivial solution, then there are no free variables, so every column of A has a pivot, so Ax = b can never have a pivot in the augmented column. Thus Ax = b has a solution no matter what b is. Thus the columns of A span Rn.

[2] (8 points) Let B and C be 4 x 4 matrices such that det B = 2 and det C = -5, and let A be the matrix
```[ 0   4   5   1 ]             (a) Determine det (3(B-3C2)T).
[ 1   0   7   1 ]
[ 1   0   2   1 ]
[ 0   1   3   0 ].            (b) Use Cramer's rule to find x2 in the equation
Ax = b, where bT = ( 5, 7, 2, 3). (This can be done
(c) Why can't you always use Cramer's rule to solve linear equations? Give an example
of a linear system of equations for which you can't use Cramer's rule, and explain
why Cramer's rule doesn't work for your example.
```
• (a) det (3(B-3C2)T) = det (3(B-3C2)) = 34(det B)-3(det C)2 = 81 x (1/8) x 25 = 2025/8.
• (b) Cramer's rule tells us x2 = (det E)/(det A), where E is the matrix
```[ 0   5   5   1 ]
[ 1   7   7   1 ]
[ 1   2   2   1 ]
[ 0   3   3   0 ].
```
Since two of the columns of E are the same, we know that E is not invertible, so det E = 0. Thus x2 = 0.
• (c) Cramer's rule applies only to systems of linear equations whose coefficient matrix is invertible. Thus it can't be used to solve 3x + 4y = 7, for example.
[3] (6 points) The nullspace of an m x n matrix A (with real number entries) is a subset of Rn; prove that it is a subspace. (Mention the three things you have to check, and then show how to check them.)

Answer: Since A0 = 0, we know 0 is in Nul A. If v and w are in Nul A, then A(v + w) = Av + Aw = 0 + 0 = 0, so Nul A is closed under vector addition. And if c is a scalar and v is in Nul A, then A(cv) = cAv = c0 = 0, so Nul A is closed under scalar multiplication. Thus Nul A is a subspace.

[4] (9 points) For this problem, you may assume that B is the reduced row echelon form of A, where A and B are the following matrices:
```[ 3  1 -2  0  1  2  1 ]        [ 1  0  -1   0   0  -2  -3 ]   (a) Find a basis for the
[ 1  1  0 -1  1  2  2 ]        [ 0  1   1   0   0   2   3 ]       column space of A.
[ 3  2 -1  1  1  8  9 ]        [ 0  0   0   1   0   4   5 ]
[ 0  2  2 -1  1  6  8 ]        [ 0  0   0   0   1   6   7 ]   (b) Find a basis for the
[ 0  3  3  3 -3  0  3 ]        [ 0  0   0   0   0   0   0 ]       null space of A.
```
• (a) From B we see that columns 1, 2, 4 and 5 of A are the pivot columns of A, thus columns 1, 2, 4 and 5 of A give a basis for Col A.
• (b) The parametric vector form of the solution to Ax = 0 is
```[ x1]       [ 1]      [ 2]      [ 3]
[ x2]       [-1]      [-2]      [-3]
[ x3]       [ 1]      [ 0]      [ 0]
[ x4] =  x3 [ 0] + x6 [-4] + x7 [-5] .
[ x5]       [ 0]      [-6]      [-7]
[ x6]       [ 0]      [ 1]      [ 0]
[ x7]       [ 0]      [ 0]      [ 1]
```
So the vectors [1, -1, 1, 0, 0, 0, 0]T, [2, -2, 0, -4, -6, 1, 0]T, and [3, -3, 0, -5, -7, 0, 1]T give a basis of Nul A.
[5] (6 points) Find the coordinates of ( 1, 2)T with respect to the basis B = {( 1, 4)T, ( 1, 1)T} of R2.

Answer: Let v = ( 1, 4)T, w = ( 1, 1)T, and u = ( 1, 2)T. The coordinates of u with respect to B are (a, b), where av + bw = u. I.e., we have to solve the system of equations av + bw = u for a and b. We get a = 1/3, and b = 2/3.

[6] (6 points): Say A is a 5x7 matrix with dim Nul(A)=4. Find the dimension of the row space of A and of the null space of AT.
ANSWER: Since dim Nul(A)=4, we know rank(A) = 7 - 4 = 3. But rank(A) = dim Row A, so dim Row A = 3. But rank(AT) = rank(A), so dim Nul(AT) = 5 - rank(AT) = 5 - 3 = 2.

[7] (6 points): Let A be the matrix
```                  [ 1  -4   3 ]
[ 4  -4   0 ]
[ 2  -4   2 ].
```
In each case determine if the given vector is an eigenvector of A and if so find its eigenvalue.
• (a) v = [1,1,1]T
ANSWER: Av = 0v, so v is an eigenvector of eigenvalue 0.
• (b) v = [1,0,1]T
ANSWER: Av = 4v, so v is an eigenvector of eigenvalue 4.
• (c) v = [0,0,0]T
ANSWER: The zero vector is never an eigenvector, so v cannot be an eigenvector.
[8] (10 points): Let A be the matrix
```                  [ 1  1  1 ]
[ 0  2  1 ]
[ 0  0  3 ].
```
This is a diagonalizable matrix; thus P-1AP=D for some invertible matrix P, where D is a diagonal matrix.
• (a) Find each eigenvalue l of A, and a basis for each eigenspace El.
ANSWER: Since A is triangular, the eigenvalues are on the diagonal: 1, 2, and 3. Since we have three different eigenvalues for a 3x3 matrix, we know each eigenspace will be 1-dimensional, so now we just need to find a single eigenvector for each eigenvalue. One way to do this is to solve (A - lI)x = 0, for each eigenvalue l. It is also sometimes possible to find them by inspection. In any case, [1, 0, 0]T works for l = 1, [1, 1, 0]T works for l = 2, and [1, 1, 1]T works for l = 3.
• (b) Find P and D.
ANSWER: D is the diagonal matrix whose diagonal entries are 1, 2 and 3, and P is the matrix whose columns are the corresponding eigenvectors, [1, 0, 0]T, [1, 1, 0]T and [1, 1, 1]T.
[9] (8 points): For each of the matrices below, determine if it is diagonalizable. Give a reason why or why not. (These can all be done by inspection.)

```(a) The 5x5 identity matrix.          (b) [ 1  1  0  0  0 ] ANSWER: This is not diagonalizable,
ANSWER: This is diagonal,                 [ 0  1  1  0  0 ] since the 0 eigenspace has algebraic
hence diagonalizable.                     [ 0  0  1  0  0 ] multiplicity 2 but dimension 1.
[ 0  0  0  0  1 ] (Also, the 1 eigenspace has algebraic
[ 0  0  0  0  0 ] multiplicity 3 but dimension 1.)

(c) A 5x5 matrix A with eigenvalues    (d) A matrix of the form PDP-1, where
0, 2, and 3, and such that the             P is invertible and D is diagonal.
dimension of the null space is 2           ANSWER: The matrix PDP-1 is diagonalizable
for both A and A-3I.                       by definition.
```                  [ 0  -1 ]