Mathematics 244, section 1 – Linear Algebra Final Examination Solutions May 10, 2007

I. Let

A = ( \begin{matrix}−1 & 0 & 2 & 2 & −1 \\ 2 & 1 & −1 & 2 & −1 \\ 1 & a & −5 & −8 & a \end{matrix} )

A) (10) Are there any a ∈ R such that columns 1,2,3 of A form a linearly dependent set? If so, find such an a. If not, say why not.

Solution: Performing row operations on the first three columns:

\begin{aligned}( \begin{matrix}−1 & 0 & 2 \\ 2 & 1 & −1 \\ 1 & a & −5 \end{matrix} ) & ∼ ( \begin{matrix}−1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & a & −3 \end{matrix} ) \\ & ∼ ( \begin{matrix}−1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & −3 − 3a \end{matrix} ) \end{aligned}

The first three columns will be linearly dependent if −3 − 3a = 0, or a = −1. They will be linearly independent for any other a ∈ R.

B) (10) Are there any a ∈ R such that the system of equations Ax = e_{1} (the first standard basis vector in R^{3}) is inconsistent? If so, for which a is this true? If not, say why not.

Solution: Do the same row operations on the augmented matrix of the system Ax = e_{1}:

( \begin{matrix}−1 & 0 & 2 & 2 & −1 & | & 1 \\ 2 & 1 & −1 & 2 & −1 & | & 0 \\ 1 & a & −5 & −8 & a & | & 0 \end{matrix} )

∼ ( \begin{matrix}−1 & 0 & 2 & 2 & −1 & | & 1 \\ 0 & 1 & 3 & 6 & −3 & | & 2 \\ 0 & a & −3 & −6 & a − 1 & | & 1 \end{matrix} )

∼ ( \begin{matrix}−1 & 0 & 2 & 2 & −1 & | & 1 \\ 0 & 1 & 3 & 6 & −3 & | & 2 \\ 0 & 0 & −3 − 3a & −6 − 6a & 4a − 1 & | & 1 − a \end{matrix} )

From this we can see that the answer to the question is No. The first two columns always contain pivots. If a ≠ −1, then there is a third pivot in column 3, so there are solutions. If a = −1, then there is a third pivot in column 5, since 4(−1)−1 = −5 ≠ 0. Therefore, the system Ax = e_{1} has solutions for all a ∈ R.

C) (15) Let a = 4 and find a parametrization for Nul(A).

Solution: Continuing from above with a = 4 (but omitting the last column):

( \begin{matrix}−1 & 0 & 2 & 2 & −1 \\ 0 & 1 & 3 & 6 & −3 \\ 0 & 0 & −15 & −30 & 15 \end{matrix} )

∼ ( \begin{matrix}1 & 0 & −2 & −2 & 1 \\ 0 & 1 & 3 & 6 & −3 \\ 0 & 0 & 1 & 2 & −1 \end{matrix} )

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

Hence

Nul(A) = Span \{( \begin{matrix}−2 \\ 0 \\ −2 \\ 1 \\ 0 \end{matrix} ) , ( \begin{matrix}1 \\ 0 \\ 1 \\ 0 \\ 1 \end{matrix} )\} = \{s ( \begin{matrix}−2 \\ 0 \\ −2 \\ 1 \\ 0 \end{matrix} ) + t ( \begin{matrix}1 \\ 0 \\ 1 \\ 0 \\ 1 \end{matrix} ) : s, t ∈ R\}

II. A) (5) Define: The span of a set S in a vector space V.

Solution: The span of S is the set of all linear combinations of vectors in S, or

Span(S) = {c_{1}x_{1} + ⋯ + c_{k}x_{k} : x_{i} ∈ S, c_{i} ∈ R}.

B) (10) Given: if V is a vector space, S is a finite subset of V with Span(S) = V, and T is any linearly independent subset of V, then |T | ≤ |S|. Use this fact to show that any two bases of V must have the same number of elements.

Solution: Let B and C be any two bases of V. Since B is linearly independent and C spans V, we have by the given fact |B| ≤ |C|. On the other hand, since B spans V and C is linearly independent, we have |B| ≥ |C|. Since both inequalities are true, |B| = |C|.

C) (5) The result of part B provides the justification for the definition of what concept connected to vector spaces?

Solution: This is the justification for the definition of the dimension of a vector space as the number of vectors in a basis.

A) (15) Solve by Cramer’s Rule:

\begin{aligned}3x_{1} + x_{3} & = 3 \\ 4x_{1} + 3x_{2} − x_{3} & = 2 \\ x_{2} + 2x_{3} & = 1 \end{aligned}

Solution: Computing determinants, we have

det(A) = 3 · 7 − 0 + 1 × 4 = 25
det(A_{1}(b)) = det ( \begin{matrix}3 & 0 & 1 \\ 2 & 3 & −1 \\ 1 & 1 & 2 \end{matrix} ) = 3 × 7 + 1(−1) = 20
det(A_{2}(b)) = det ( \begin{matrix}3 & 3 & 1 \\ 4 & 2 & −1 \\ 0 & 1 & 2 \end{matrix} ) = 3 × 5 − 3 × 8 + 1 × 4 = −5
det(A_{1}(b)) = det ( \begin{matrix}3 & 0 & 3 \\ 4 & 3 & 2 \\ 0 & 1 & 1 \end{matrix} ) = 3 × 1 + 0 + 3 × 4 = 15

Therefore,

x_{1} = \frac{det(A_{1}(b))}{det(A)} = \frac{20}{25} = \frac{4}{5}
x_{2} = \frac{det(A_{2}(b))}{det(A)} = \frac{−5}{25} = −\frac{1}{5}
x_{3} = \frac{det(A_{3}(b))}{det(A)} = \frac{15}{25} = \frac{3}{5}.

(It is easy to check that these are correct, by substituting in the original equations.) B) (5) Suppose det(A) = 3, det(B) = −2, and det(C) = 4. What is det(A^{t}B^{−1}C^{2})?

Solution: By the properties of determinants, det(A^{t}B^{−1}C^{2}) = det(A) det(B)^{−1}(det(C))^{2} = 3 × \frac{1}{(−2)} × 4^{2} = −24.

IV. All parts of this problem refer to the linear mapping T : P_{3}(R) → R^{3} defined by

T (p) = ( \begin{matrix}p(−1) \\ p(0) \\ p(1) \end{matrix} )

A) (10) Show that the image of T is a vector subspace of R^{3} directly (that is, do not just quote a general theorem).

Solution: The image Im(T) is the set of y ∈ R^{3} such that y = T (p) for some polynomial p ∈ P_{3}(R). Since T (0) = 0, we know 0 ∈ Im(T). Let y_{1}, y_{2} ∈ Im(T). Then y_{1} = T (p_{1}) and y_{2} = T (p_{2}) for some p_{1}, p_{2}. But then since T is linear:

y_{1} + y_{2} = T (p_{1}) + T (p_{2}) = T (p_{1} + p_{2}).

This shows y_{1} + y_{2} ∈ Im(T) also. Finally if y = T (p) ∈ Im(T) and c ∈ R, then cy = cT (p) = T (cp). Hence cy ∈ Im(T). Therefore, Im(T) is a subspace of R^{3}.

B) (10) Find the matrix of T with respect to the basis B = {1, x, x^{2}, x^{3}} in the domain and the standard basis in R^{3}.

Solution: We have

T (1) = ( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) , T (x) = ( \begin{matrix}−1 \\ 0 \\ 1 \end{matrix} ) , T (x^{2}) = ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ) , and T (x^{3}) = T (x).

Therefore, by definition:

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

C) (15) State the change of coordinates formula and use it to find the matrix of T with respect to the basis B in the domain and the basis

C = \{( \begin{matrix}1 \\ 1 \\ 0 \end{matrix} ) , ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ) , ( \begin{matrix}0 \\ 1 \\ 1 \end{matrix} )\}

in R^{3}. For full credit, you must show all details of the computation of any matrix inverses you need.

Solution: The change of coordinates formula says that if T : V → W is linear, B, B′ are bases of V, and CC′ are bases of W, then

[T]_{C′←B′} = P_{C′←C}[T]_{C←B}[T]_{B←B′}.

Here we want:

[T]_{C←B} = P_{C←std}[T]_{std←B}

= (P_{std←C})^{−1}[T]_{std←B}

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

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

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

The inverse matrix is computed as usual by row reducing (P |I) ∼ (I|P^{−1}); details omitted.

V. (10) Show by mathematical induction: If A = Q^{−1}BQ for some invertible Q, then for all integers n ≥ 1, A^{n} = Q^{−1}B^{n}Q.

Solution: The base case of the induction is n = 1, and there is nothing to prove. For the induction step, assume A^{k} = Q^{−1}B^{k}Q. Then using the induction hypothesis,

A^{k+1} = A · A^{k} = (Q^{−1}BQ)(Q^{−1}B^{k}Q)
= Q^{−1}B(QQ^{−1})B^{k}Q = Q^{−1}BB^{k}Q = Q^{−1}B^{k+1}Q,

which is what we wanted to show.

VI. Let A be the symmetric matrix:

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

where a, b are real numbers with b > a > 0. A) (10) Show that A has eigenvalues λ = a + b, a, a − b and determine the corresponding

eigenspaces.

Solution: We have

det A − λI = det ( \begin{matrix}a − λ & 0 & b \\ 0 & a − λ & 0 \\ b & 0 & a − λ \end{matrix} ) .

Expanding along row 2, we have

= (a − λ) det ( \begin{matrix}a − λ & b \\ b & a − λ \end{matrix} ) = (a − λ)(λ^{2} − 2aλ + (a^{2} − b^{2})).

This factors easily as

= (a − λ)((a + b) − λ)((a − b) − λ).

Hence the eigenvalues are λ = a, a + b, a − b. Under the assumption b > a > 0, all three eigenvalues are distinct. We have

Nul(A − aI) = Nul ( \begin{matrix}0 & 0 & b \\ 0 & 0 & 0 \\ b & 0 & 0 \end{matrix} ) = Span \{( \begin{matrix}0 \\ 1 \\ 0 \end{matrix} )\}

Nul(A − (a + b)I) = Nul ( \begin{matrix}−b & 0 & b \\ 0 & −b & 0 \\ b & 0 & −b \end{matrix} ) = Span \{( \begin{matrix}1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{matrix} )\}

Nul(A − (a − b)I) = Nul ( \begin{matrix}b & 0 & b \\ 0 & b & 0 \\ b & 0 & b \end{matrix} ) = Span \{( \begin{matrix}−1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{matrix} )\}

B) (10) Find the spectral decomposition of A.

Solution: Write u_{i} for the basis vectors above for the three eigenspaces. Note that we normalized so that {u_{1}, u_{2}, u_{3}} is an orthonormal basis of R^{3} (this is necessary for the next step). The spectral decomposition is A = au_{1}u_{1}^{t} + (a + b)u_{2}u_{2}^{t} + (a − b)u_{3}u_{3}^{t} =

a · ( \begin{matrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} ) + (a + b) · ( \begin{matrix}1/2 & 0 & 1/2 \\ 0 & 0 & 0 \\ 1/2 & 0 & 1/2 \end{matrix} ) + (a − b) · ( \begin{matrix}1/2 & 0 & −1/2 \\ 0 & 0 & 0 \\ −1/2 & 0 & 1/2 \end{matrix} ) .

VII. True - False. For each true statement give a brief proof (saying “this is true by definition” is not sufficient!). For each false statement give a counterexample or a reason.

A) (10) If A is an n × n matrix and AQ = QD for some diagonal matrix D and some matrix Q, then A is diagonalizable.

Solution: This is False. The equation AQ = QD does mean that all the columns of Q are eigenvectors of A. But it is not stated that Q is invertible, which means that the columns of Q do not need to form a basis for R^{n}, and A does not need to be diagonalizable. A counterexample: let A = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) which is not diagonalizable. If

Q = ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ), then

AQ = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ) I_{2} = QD,

where D = I_{2} is the identity matrix.

B) (10) If A ∈ M_{4×5}(R) and dim Col(A) = 2, then there are 4 free variables in the system Ax = 0.

Solution: This is False. A reason is that dim Nul(A)+dim Col(A) = 5, so if dim Col(A) = 2, then dim Nul(A) = 3. Since the set of solutions of Ax = 0 is Nul(A) by definition, there will only be 3 free variables here.

C) (10) If {v_{1}, v_{2}, v_{3}} is a linearly independent subset of R^{3} and V is the matrix with columns v_{1}, v_{2}, v_{3}, then det(V) ≠ 0.

Solution: This is True. By one of the parts of the Invertible Matrix Theorem (see p. 129 in the text), V is invertible. Hence det(V) ≠ 0.

D) (10) If A is a symmetric matrix and v_{1}, v_{2}, …, v_{k} are eigenvectors of A with distinct corresponding eigenvalues, then {v_{1}, v_{2}, …, v_{k}} is an orthogonal set.

Solution: This is True. Suppose Av_{i} = λ_{i}v_{i} with distinct λ_{i}. Then for all pairs i ≠ j,

λ_{i}(v_{i} · v_{j}) = (λ_{i}v_{i}) · v_{j}
= (Av_{i}) · v_{j}
= v_{i} · (Av_{j}) A symmetric
= v_{i} · (λ_{j}v_{j})
= λ_{j}(v_{i} · v_{j})

This shows 0 = (λ_{i} − λ_{j})(v_{i} · v_{j}). Since the first factor is not zero, the second one must be. Therefore v_{i} and v_{j} are orthogonal. Since this is true for all pairs i, j, the set of eigenvectors is an orthogonal set.

VIII. (The Cayley-Hamilton Theorem.)

A) (10) Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) be any 2 × 2 matrix. Show that A “satisfies its own character-

istic polynomial equation” in the sense that A^{2} − Tr(A)A + det(A)I_{2} = 0_{2×2} (here I_{2} is the 2 × 2 identity matrix and 0_{2×2} is the 2 × 2 zero matrix).

Solution: Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ). By direct calculation: A^{2} − Tr(A)A + det(A)I_{2} =

( \begin{matrix}a^{2} + bc & ab + bd \\ ca + dc & cb + d^{2} \end{matrix} ) − (a + d) ( \begin{matrix}a & b \\ c & d \end{matrix} ) + ( \begin{matrix}ad − bc & 0 \\ 0 & ad − bc \end{matrix} )

= ( \begin{matrix}a^{2} + bc − a^{2} − ad + ad − bc & ab + bd − ab − bd \\ ca + dc − ac − dc & cb + d^{2} − ad − d^{2} + ad − bc \end{matrix} )

= ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) .

Thus the Cayley-Hamilton theorem is true for 2 × 2 matrices.

B) (10) Prove that an n × n diagonalizable matrix A satisfies its own characteristic polynomial in the same sense as in part A: If

det(A − tI) = (−1)^{n}t^{n} + c_{n−1}t^{n−1} + ⋯ + c_{1}t + c_{0},

then

(−1)^{n}A^{n} + c_{n−1}A^{n−1} + ⋯ + c_{1}A + c_{0}I_{n} = 0_{n×n}.

Caution: It does not make sense to “substitute t = A” in det(A − tI)” because t is a scalar, not a matrix. The result of problem V above may be useful. (Comment: The result is also true without the extra hypothesis of diagonalizability but it is more difficult to prove then.)

Solution: Under the assumption that A is diagonalizable, there is an invertible matrix Q and a diagonal matrix D with diagonal entries λ_{i}, the eigenvalues of A, such that A = Q^{−1}DQ. By problem V above, for all positive integers k, A^{k} = Q^{−1}D^{k}Q, and the diagonal entries of D^{k} are just the powers of the eigenvalues λ_{i}^{k}. Substituting for the various powers of A, and factoring out Q^{−1} on left and Q on right,

(−1)^{n}A^{n} + c_{n−1}A^{n−1} + ⋯ + c_{1}A + c_{0}I_{n}
= (−1)^{n}Q^{−1}D^{n}Q + c_{n−1}Q^{−1}D^{n−1}Q + ⋯ + c_{1}Q^{−1}AQ + c_{0}Q^{−1}Q
= Q^{−1}((−1)^{n}D^{n} + c_{n−1}D^{n−1} + ⋯ + c_{1}D + c_{0}I_{n})Q

Since D is diagonal, the middle matrix here,

(−1)^{n}D^{n} + c_{n−1}D^{n−1} + ⋯ + c_{1}D + c_{0}I_{n},

is also a diagonal matrix. The ith diagonal entry is

(−1)^{n}λ_{i}^{n} + c_{n−1}λ_{i}^{n−1} + ⋯ + c_{1}λ_{i} + c_{0}.

For all i, this equals 0, since it is det(A − λ_{i}I) = 0. Hence the whole matrix in the middle is the n × n zero matrix, which means the whole product is the zero matrix.

IX. (Eigenvalues in calculus!) Let V = C^{∞}(R) be the vector space of all functions f : R → R that have derivatives of all orders at all x ∈ R.

A) (10) Let D : V → V be the linear mapping defined by D(f) = f′. Show that for all λ ∈ R, the eigenspace Ker(D − λI) has dimension 1, and find a basis.

Solution: We have that Ker(D − λI) equals the set of all functions f satisfying (D − λI)(f) = 0. Here I means the identity mapping on V, so I(f) = f for all f. We want to find all functions satisfying:

D(f) = f′ = λf, or \frac{df}{dx} = λf.

This is just the usual exponential growth/decay differential equation. We can solve it by separation of variables as usual:

∫ \frac{df}{f} = ∫ λdx
ln |f| = λx + k
f = ±e^{k}e^{λx}
= ce^{λx},

where c ∈ R is arbitary. Hence the eigenspace is one-dimensional, and {e^{λx}} is one basis.

B) (5) Let S : V → V be the linear mapping defined by S(f) = f′′. Let λ = −k^{2} for an arbitrary k > 0. Find two linearly independent functions in the eigenspace Ker(S − λI).

Solution: Recall the derivative formulas

\frac{d}{dx} sin(x) = cos(x); \frac{d}{dx} cos(x) = − sin(x).

Thus

S(sin(x)) = \frac{d^{2}}{dx^{2}} sin(x) = − sin(x); S(cos(x)) = \frac{d^{2}}{dx^{2}} cos(x) = − cos(x).

By the chain rule for derivatives,

{sin(kx), cos(kx)}

satisfy

S(sin(kx)) = \frac{d^{2}}{dx^{2}} sin(kx) = −k^{2} sin(kx); S(cos(kx)) = \frac{d^{2}}{dx^{2}} cos(kx) = −k^{2} cos(kx).

So {sin(kx), cos(kx)} is one such set of functions.

C) (5) Let S : V → V be the linear mapping defined by

S(f) = f(1) + f′(1)(x − 1) + \frac{f′′(1)}{2}(x − 1)^{2}.

Show that Span{1, x, x^{2}} ⊆ Ker(S − I). For f(x) = x^{2},

T (f) = f(1) + f′(1)(x − 1) + \frac{f′′(1)}{2}(x − 1)^{2}

= 1 + 2 · (x − 1) + \frac{2}{2}(x − 1)^{2}

= 1 + 2x − 2 + x^{2} − 2x + 1

= x^{2}

Thus T (f) = f and f is an eigenfunction of T with eigenvalue 1. The proofs for f = 1, x are similar, and easier. Since Ker(S − I) is a subspace of V, this shows Span{1, x, x^{2}} ⊆ Ker(S − I). (Note what this says: the second degree Taylor polynomial of a polynomial of degree ≤ 2 at a = 1 equals the polynomial itself. The same thing is true for all a, and all degrees! This also shows that the inclusion is an equality!)