Mathematics 244, section 1 – Linear Algebra Exam 3 Solutions – April 30, 2007

I. Let V = Span{e^{x}, cos(x), sin(x)}, let W = P_{2}(R) and let T : V → W be the mapping defined by

T (f) = f(0) + f′(0)x + \frac{f′′(0)}{2}x^{2}.

A) (10) Show that T is a linear mapping.

Solution: Let f, g ∈ V. Then

T (f + g) = (f + g)(0) + (f + g)′(0)x + \frac{(f + g)′′(0)}{2}x^{2}

= f(0) + f′(0)x + \frac{f′′(0)}{2}x^{2} + g(0) + g′(0)x + \frac{g′′(0)}{2}x^{2}

= T (f) + T (g).

Similarly, for c ∈ R,

T (cf) = (cf)(0) + (cf)′(0)x + \frac{(cf)′′(0)}{2}x^{2}

= cf(0) + cf′(0)x + c\frac{f′′(0)}{2}x^{2}

= cT (f).

B) (15) Let B = {e^{x}, cos(x), sin(x)} (a basis for V) and C = {1, x, x^{2}} (basis for W). Find the matrix [T]_{C←B}.

Solution: Note that T (f) is just the second-degree Taylor polynomial of f at a = 0. So we can compute T (f) using facts from calculus:

\begin{aligned}T (e^{x}) & = 1 + x + \frac{x^{2}}{2} = 1 · 1 + 1 · x + \frac{1}{2} · x^{2} \\ T (cos(x)) & = 1 − \frac{x^{2}}{2} = 1 · 1 + 0 · x + \frac{−1}{2} · x^{2} \\ T (sin(x)) & = x = 0 · 1 + 1 · x + 0 · x^{2} \end{aligned}

Hence placing the coordinate vectors of these T (f) into the columns,

[T]_{C←B} = ( \begin{matrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ \frac{1}{2} & \frac{−1}{2} & 0 \end{matrix} ) .

C) (15) Now, let C′ = {x^{2} + x, x − 2, x + 1}. Find the matrix [T]_{C′←B}.

Solution: We have

[T]_{C′←B} = P_{C′←C}[T]_{C←B}.

The matrix

P_{C′←C} = (P_{C←C′})^{−1} = ( \begin{matrix}0 & −2 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} )^{−1} .

Computing the inverse,

( \begin{matrix}0 & −2 & 1 & | & 1 & 0 & 0 \\ 1 & 1 & 1 & | & 0 & 1 & 0 \\ 1 & 0 & 0 & | & 0 & 0 & 1 \end{matrix} ) ∼ ( \begin{matrix}1 & 0 & 0 & | & 0 & 0 & 1 \\ 0 & −2 & 1 & | & 1 & 0 & 0 \\ 1 & 1 & 1 & | & 0 & 1 & 0 \end{matrix} )

∼ ( \begin{matrix}1 & 0 & 0 & | & 0 & 0 & 1 \\ 0 & −2 & 1 & | & 1 & 0 & 0 \\ 0 & 1 & 1 & | & 0 & 1 & −1 \end{matrix} )

∼ ( \begin{matrix}1 & 0 & 0 & | & 0 & 0 & 1 \\ 0 & 1 & −1/2 & | & −1/2 & 0 & 0 \\ 0 & 1 & 1 & | & 0 & 1 & −1 \end{matrix} )

∼ ( \begin{matrix}1 & 0 & 0 & | & 0 & 0 & 1 \\ 0 & 1 & −1/2 & | & −1/2 & 0 & 0 \\ 0 & 0 & 3/2 & | & 1/2 & 1 & −1 \end{matrix} )

∼ ( \begin{matrix}1 & 0 & 0 & | & 0 & 0 & 1 \\ 0 & 1 & 0 & | & 1/3 & 1/3 & −1/3 \\ 0 & 0 & 1 & | & 1/3 & 2/3 & −2/3 \end{matrix} )

So

( \begin{matrix}0 & −2 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} )^{−1} = ( \begin{matrix}0 & 0 & 1 \\ −1/3 & 1/3 & −1/3 \\ 1/3 & 2/3 & −2/3 \end{matrix} ) .

Then

[T]_{C′←B} = ( \begin{matrix}0 & 0 & 1 \\ −1/3 & 1/3 & −1/3 \\ 1/3 & 2/3 & −2/3 \end{matrix} ) ( \begin{matrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1/2 & −1/2 & 0 \end{matrix} ) = ( \begin{matrix}1/2 & −1/2 & 0 \\ −1/6 & −1/6 & 1/3 \\ 2/3 & 2/3 & 2/3 \end{matrix} ) .

II. Let

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

The characteristic polynomial of A is p(λ) = −λ^{3} + (s + 2)λ^{2} − (5s + 1)λ + 4s.

A) (10) Determine a value of s such that λ = 3 will be an eigenvalue of A.

Solution: This will be the case when

0 = −27 + 9(s + 2) − 3(5s + 1) + 4s = −2s − 12,

so s = −6.

B) (20) Let s = −1. Is A diagonalizable in this case? If so, find an invertible matrix Q and a diagonal matrix D such that Q^{−1}AQ = D. If not, say why not. Note: You may “purchase” a factored form of the characteristic polynomial with s = −1 from me during the exam for a “price” of 5 points on this part.

Solution: When s = −1, the characteristic polynomial is

−λ^{3} + λ^{2} + 4λ − 4 = (λ − 1)(4 − λ^{2}) = (λ − 1)(2 − λ)(2 + λ).

The eigenvalues are λ = 1, 2, −2. These are distinct, so we know A is diagonalizable. (Note this also follows from the Spectral Theorem, since A is symmetric when s = −1(!)) We have for λ = 1,

Nul(A − I) = Nul ( \begin{matrix}0 & −1 & −1 \\ −1 & 2 & −1 \\ −1 & −1 & 0 \end{matrix} ) = Span \{( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} )\} .

For λ = 2,

Nul(A − 2I) = Nul ( \begin{matrix}−1 & −1 & −1 \\ −1 & 3 & −1 \\ −1 & −1 & −1 \end{matrix} ) = Span \{( \begin{matrix}−1 \\ 0 \\ 1 \end{matrix} )\} .

Finally, for λ = −2,

Nul(A + 2I) = Nul ( \begin{matrix}3 & −1 & −1 \\ −1 & 1 & −1 \\ −1 & −1 & 3 \end{matrix} ) = Span \{( \begin{matrix}1 \\ 2 \\ 1 \end{matrix} )\} .

So

D = ( \begin{matrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & −2 \end{matrix} ) and Q = ( \begin{matrix}1 & −1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 1 \end{matrix} ) .

III. (True/False) For each true statement, give a short proof. For each false statement, give a counterexample. Do any three parts. If you submit solutions for all four, I will count any additional points you earn as extra credit.

A) (10) Let A be an n×n matrix. If {v_{1}, …, v_{k}} are eigenvectors with distinct eigenvalues λ_{i}, i = 1, …, k, then {v_{1}, …, v_{k}} is an orthogonal set.

Solution: This is False. A counterexample: Let A = ( \begin{matrix}1 & 1 \\ 0 & 2 \end{matrix} ). Then λ = 1, 2 are

eigenvalues, with corresponding eigenvectors v_{1} = ( \begin{matrix}1 \\ 0 \end{matrix} ) and v_{2} = ( \begin{matrix}1 \\ 1 \end{matrix} ). But v_{1} · v_{2} =

1 ≠ 0. (The statement is true if A is symmetric, though.)

B) (10) If A and B are n × n matrices with the same characteristic polynomial, then A and B are similar.

Solution: This is False. A counterexample: Let A = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) and B = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ). Both

A and B have characteristic polynomial (λ − 1)^{2}. But A is not diagonalizable while B is diagonalizable (diagonal, even). Hence they cannot be similar. (The converse of this statement is true.)

C) (10) If v is a nonzero vector in R^{n}, and P_{v} is the orthogonal projection mapping to the line spanned by v, then the matrix [P_{v}]_{B←B} has rank 1 no matter which basis B is used in R^{n}.

Solution: This is True. P_{v}(w) = \frac{v·w}{v·v} v is a multiple of v for all w. Hence if B = {b_{1}, …, b_{n}} is any basis, then for all i, P_{v}(b_{i}) are scalar multiples of v. Hence [P_{v}(b_{i})]_{B} will all be scalar multiples of [v]_{B}. It follows that the dimension of the column space of [P_{v}]_{B←B} is 1 for all bases B.

D) (10) If a 6×6 matrix A has characteristic polynomial (1−λ)^{3}(6−λ)^{3}, and rank(A−I) = 4 and rank(A − 6I) = 3, then A is diagonalizable.

Solution: This is False. Since rank(A − I) = 4, dim Nul(A − I) = 2. Similarly, since rank(A − 6I) = 3, dim Nul(A − 6I) = 3. Since 2 + 3 < 6, A cannot be diagonalizable.