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
in your head, but show how you obtain your answer.)
(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.
Answer:
[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.
Answer:
[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.
ANSWER: Since the algebraic multiplicity
and dimension of each eigenspace are equal, we know this matrix is diagonalizable.
[10] (8 points): Let A be the matrix
[ 0 -1 ]
[ 2 3 ].
- (a) Find the eigenvalues of A.
ANSWER: From the characteristic equation we find the
eigenvalues to be 1 and 2.
- (b) Find a basis for the eigenspace corresponding
to each eigenvalue of A.
ANSWER: A basis for the 1 eigenspace is given by [1, -1]T and for
the 2 eigenspace a basis is given by [1, -2]T.
- (c) Find A10v, where v =
[1, 0]T.
ANSWER: Since [1, 0]T = 2[1, -1]T - [1, -2]T,
A10[1, 0]T = 2(110)[1, -1]T -
210[1, -2]T = [ -1022, 2046]T.