Mathematics 244, section 1 – Linear Algebra Retest Exam 2 – March 28, 2007

I. All parts of this question refer to the matrix

A = ( \begin{matrix}1 & 4 & 2 & 1 & 2t \\ −3 & 2 & 2 & 3 & t^{2} \\ 2 & 1 & 0 & −1 & −t \end{matrix} )

A) (15) What is dim Nul(A) if t = 2? Find a basis for Nul(A) in this case.

Solution: The row-reduced echelon form with t = 2 is

E = ( \begin{matrix}1 & 0 & −2/7 & −5/7 & 0 \\ 0 & 1 & 4/7 & 3/7 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} )

This shows Ax = 0 has two free variables, so dim Nul(A) = 2. A basis is

\{( \begin{matrix}2/7 \\ −4/7 \\ 1 \\ 0 \\ 0 \end{matrix} ) , ( \begin{matrix}5/7 \\ −3/7 \\ 0 \\ 1 \\ 0 \end{matrix} )\}

B) (10) What is dim Col(A) if t = 4. Find a basis for Col(A) in this case.

Solution: With t = 4, the row-reduced echelon form has a row of zeroes, and pivots in columns 1,2. A basis for Col(A) consists of the first two columns of A.

II. A) (15) Give the general proof of Cramer’s Rule for systems Ax = b with A ∈ M_{n×n} an

invertible matrix.

Solution: See the class notes or the proof of Theorem 7 in section 3.3 of the text.

B) (10) Give an example of a system Ax = b, where A is an invertible 3 × 3 matrix, but the second component of the solution vector x is x_{2} = 0.

Solution: Here’s the way I was thinking of this one. By Cramer’s rule, x_{2} = \frac{det(A_{2}(b))}{det(A)}. So one way to get x_{2} = 0 is to make b a linear combination of columns 1 and 3 of A. Then det(A_{2}(b)) = 0. An example is Ax = b where

A = ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} ) , b = ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} )

Note that det(A_{2}(b)) will have two equal columns in this case, so its determinant is automatically zero.

III. Let V = P_{2}(R), the vector space of polynomials of degree ≤ 2, and let W = M_{2×2}(R), the vector space of all 2 × 2 matrices with real entries.

A) (10) Show that T : V → W defined by T (p) = ( \begin{matrix}p(1) & p(2) \\ p′(1) & p′(2) \end{matrix} ) (p′ = derivative of p)

is a linear mapping.

Solution: If p, q ∈ V, then

T (p+q) = ( \begin{matrix}(p + q)(1) & (p + q)(2) \\ (p + q)′(1) & (p + q)′(2) \end{matrix} ) = ( \begin{matrix}p(1) + q(1) & p(2) + q(1) \\ p′(1) + q′(1) & p′(2) + q′(2) \end{matrix} ) = T (p)+T (q).

Similarly, if r ∈ R,

T (rp) = ( \begin{matrix}(rp)(1) & (rp)(2) \\ (rp)′(1) & (rp)′(2) \end{matrix} ) = r ( \begin{matrix}p(1) & p(2) \\ p′(1) & p′(2) \end{matrix} ) = rT (p).

Hence T is linear.

B) (10) Let K = \{A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ W | dim Col(A) ≤ 1\} Is K a vector subspace of W?

Why or why not?

Solution: The answer is No. K is not closed under matrix sums. The condition dim Col(A) ≤ 1 means that the columns of A are scalar multiples (possibly both the zero vector). If we take

A = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) , B = ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} )

Then A, B ∈ W since each has column space of dimension 1. But A + B = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

which is not in K, since its column space is R^{2}. (The other two properties for subspaces do hold in this case.)

C) (5 Extra Credit) Let T be the mapping from part A. What is Ker(T)? Justify your answer.

Solution: Ker(T) is the set of polynomials p = ax^{2} + bx + c ∈ P_{2}(R) such that

T (p) = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ), that is, p(1) = p(2) = 0 and p′(1) = p′(2) = 0. From the equation

on the derivatives we see that y = p(x) must have two critical points at x = 1, 2. This implies a = 0 since a ≠ 0 implies p′(x) = 2ax+b = 0 at only one point x = −b/(2a) (an “honest-to-God” parabola has only one turning point). But then the graph y = p(x) is a straight line. The conditions p(1) = p(2) say that line passes through (1, 0) and (2, 0). Hence b = c = 0 as well. The kernel contains only the zero polynomial.

IV. (True-False) For each true statement, give a short proof. For each false statement, give a counterexample. (Note: As always, “true” means “true in every case.” Partial credit will be given for definitions or theorems that apply even if you don’t find a complete solution.)

A) (10) If a, b, c are distinct real numbers, then the matrix V = ( \begin{matrix}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} ) is invertible.

Solution: TRUE. Perform row operations to compute det(V):

det ( \begin{matrix}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} ) = det ( \begin{matrix}1 & a & a^{2} \\ 0 & b − a & b^{2} − a^{2} \\ 0 & c − a & c^{2} − a^{2} \end{matrix} ) .

We can pull out factors of b − a from row 2 and c − a from row 3:

det(V) = (b − a)(c − a) det ( \begin{matrix}1 & a & a^{2} \\ 0 & 1 & b + a \\ 0 & 1 & c + a \end{matrix} ) = (b − a)(c − a)(c − b).

This is zero only when a = b, a = c, or b = c. If a, b, c are distinct, then det(V) ≠ 0 and V is invertible.

B) (10) If A is a 5 × 7 matrix, then dim Nul(A) · dim Col(A) can equal 9.

Solution: FALSE. dim Nul(A) + dim Col(A) = 7 and dim Col(A) ≤ 5. None of the possible pairs of numbers adding up to 7 have product = 9.

C) (10) Let

A = ( \begin{matrix}1 & −2 & 0 \\ −2 & −1 & −1 \end{matrix} )

Any 3 × 2 matrix X such that AX = 0_{2×2} has linearly dependent columns.

Solution: TRUE. If AX = 0, the columns of X are in Nul(A). But note that the echelon form of A will have two pivot columns so dim Nul(A) = 1. A set of two vectors in a one-dimensional space must be linearly dependent (by the second “theorem to know”).