Mathematics 244, section 1 – Linear Algebra Solutions for Final Exam Review Questions May 4, 2007

I. The following matrix is the augmented matrix for a system of 3 linear equations in 4 unknowns. For which value(s) of a ∈ R does the system have a solution? Find all solutions for those a.

( \begin{matrix}−1 & 2 & 0 & 1 & | & 2 \\ 3 & 1 & 3 & 0 & | & 0 \\ 1 & 12 & 6 & 5 & | & a \end{matrix} )

Solution: Applying row operations,

\begin{aligned}( \begin{matrix}−1 & 2 & 0 & 1 & | & 2 \\ 3 & 1 & 3 & 0 & | & 0 \\ 1 & 12 & 6 & 5 & | & a \end{matrix} ) & ∼ ( \begin{matrix}1 & −2 & 0 & −1 & | & −2 \\ 0 & 7 & 3 & 3 & | & 6 \\ 0 & 14 & 6 & 6 & | & a + 2 \end{matrix} ) \\ & ∼ ( \begin{matrix}1 & −2 & 0 & −1 & | & −2 \\ 0 & 7 & 3 & 3 & | & 6 \\ 0 & 0 & 0 & 0 & | & a − 10 \end{matrix} ) \end{aligned}

The system is inconsistent unless a = 10. When a = 10, we continue reducing to row-reduced echelon form:

∼ ( \begin{matrix}1 & 0 & 6/7 & −1/7 & | & −2/7 \\ 0 & 1 & 3/7 & 3/7 & | & 6/7 \\ 0 & 0 & 0 & 0 & | & 0 \end{matrix} )

Here, x_{3} and x_{4} are free variables. The solutions in parametric form:

( \begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} ) = ( \begin{matrix}−2/7 \\ 6/7 \\ 0 \\ 0 \end{matrix} ) + x_{3} ( \begin{matrix}−6/7 \\ −3/7 \\ 1 \\ 0 \end{matrix} ) + x_{4} ( \begin{matrix}1/7 \\ −3/7 \\ 0 \\ 1 \end{matrix} ) ,

where x_{3} and x_{4} are arbitrary real numbers.

II. Let V = P_{4}(R), and let W = {p(x) ∈ V : p′(1) = 0 and p(0) = 3p(2)}. A) Show that W is a vector subspace of V.

Solution: Let p, q ∈ W, so p′(1) = q′(1) = 0 and p(0) = 3p(2), q(0) = 3q(2). Then

(p + q)′(1) = p′(1) + q′(1) = 0 + 0 = 0,

and

(p + q)(0) = p(0) + q(0) = 3p(2) + 3q(2) = 3(p + q)(2).

Hence W is closed under sums. In addition, if r ∈ R,

(rp)′(1) = rp′(1) = r · 0 = 0

and

(rp)(0) = r · p(0) = r · 3p(2) = 3(rp)(2).

Hence W is also closed under scalar multiples, and a subspace of V.

B) Find a subset S ⊂ W such that W = Span(S).

Solution: Write p ∈ V as p = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4}. Then the equations p′(1) = 0 and p(0) = 3p(2) say

\begin{aligned}0 & = a_{1} + 2a_{2} + 3a_{3} + 4a_{4} \\ 0 & = 2a_{0} + 6a_{1} + 12a_{2} + 24a_{3} + 48a_{4} \end{aligned}

In matrix form this is the homogenenous system Aa = 0 for

A = ( \begin{matrix}0 & 1 & 2 & 3 & 4 \\ 2 & 6 & 12 & 24 & 48 \end{matrix} )

(and a = the vector of coefficients – the coordinate vector of p with respect to the standard basis in V). The row-reduced echelon form is

E = ( \begin{matrix}1 & 0 & 0 & 3 & 12 \\ 0 & 1 & 2 & 3 & 4 \end{matrix} )

There are three free variables, so the dimension of the space of solutions is three.

S = Span{−2x + x^{2}, −3 − 3x + x^{3}, −12 − 4x + x^{4}}

is a set that spans W. (Note: these are the polynomial forms of the coordinate vectors spanning the solutions of the echelon form system Ea = 0.)

C) Is your set S linearly independent? Justify your assertion.

Solution: Yes it is linearly independent. If

0 = c_{1}(−2x + x^{2}) + c_{2}(−3 − 3x + x^{3}) + c_{3}(−12 − 4x + x^{4}),

then equating coefficients of x^{2} on both sides, c_{1} = 0. Similarly c_{2} = c_{3} = 0 from the coefficients of x^{3} and x^{4}. (Recall, this is a general property of the spanning set of the null space of a matrix computed from the row-reduced echelon form.)

III. Let V be a vector space. Show that a set S ⊂ V is linearly dependent if and only if there is some x ∈ S such that x is a linear combination of the vectors in Span(S − {x}).

Solution: See class notes. IV. All parts of this problem refer to the linear mapping T : R^{4} → R^{3} defined by

T (x_{1}, x_{2}, x_{3}, x_{4}) = (4x_{1} − 3x_{2} + 2x_{3} − x_{4}, x_{2} + x_{3}, x_{1} − x_{4})

(Added note: all vectors here are written as row vectors to save space, but you should read them as column vectors – that is, transpose them – to perform the calculations in the following questions.) A) Find the matrix of T with respect to the standard bases in the domain and target.

Solution: By the standard recipe T (e_{1}) = T (1, 0, 0, 0) = (4, 0, 1)

T (e_{2}) = T (0, 1, 0, 0) = (−3, 1, 0)
T (e_{3}) = T (0, 0, 1, 0) = (2, 1, 0)
T (e_{4}) = T (0, 0, 0, 1) = (−1, 0, −1)

Hence

[T]_{std←std} = ( \begin{matrix}4 & −3 & 2 & −1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & −1 \end{matrix} ) .

B) Find the matrix of T with respect to the basis

B = {(1, −1, 0, 0), (1, 1, 0, 0), (0, 0, 2, 3), (0, 0, −1, 2)}

in the domain and

C = {(0, 0, 1), (0, −1, 1), (−1, 1, 0)}

in the target. Solution: By the change of coordinates formula

[T]_{C←B} = P_{C←std}[T]_{std←std}P_{std←B}

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

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

= ( \begin{matrix}7 & 3 & 0 & −7 \\ −6 & −2 & −3 & 5 \\ −7 & −1 & −1 & 4 \end{matrix} ) .

V. Show by mathematical induction that the determinant of an n × n upper-triangular matrix A = (a_{ij}) is the product of the diagonal entries: det(A) = a_{11}a_{22} ⋯ a_{nn}.

Solution: The base case is trivial: For a 1 × 1 matrix

det(a_{11}) = a_{11}

by definition. Now assume that the result is true for all k×k upper-triangular matrices, and consider a (k+1)×(k+1) upper-triangular matrix A. Expanding along row k+1, we see

det(A) = 0 + ⋯ + 0 + (−1)^{2k+2}a_{k+1,k+1} det(A_{k+1,k+1}).

The matrix A_{k+1,k+1} is an upper-triangular k ×k matrix. So the induction hypothesis shows that

det(A_{k+1,k+1}) = a_{11} ⋯ a_{kk}.

Hence

det(A) = a_{k+1,k+1}a_{11} ⋯ a_{kk} = a_{11} ⋯ a_{kk}a_{k+1,k+1},

which is what we wanted to show.

VI. Let

A = ( \begin{matrix}2 & 9 & 0 \\ 1 & 2 & −1 \\ 0 & 0 & 5 \end{matrix} )

A) Find the eigenvalues of A.

Solution: The characteristic polynomial is

det(A − λI) = det ( \begin{matrix}2 − λ & 9 & 0 \\ 1 & 2 − λ & −1 \\ 0 & 0 & 5 − λ \end{matrix} )

Expanding along the third row

= (5 − λ) det ( \begin{matrix}2 − λ & 9 \\ 1 & 2 − λ \end{matrix} ) = (5 − λ)(λ^{2} − 4λ − 5) = (5 − λ)^{2}(−1 − λ)

The eigenvalues are λ = 5 (multiplicity 2), and λ = −1. B) For each eigenvalue λ, find a basis of Nul(A − λI).

Solution: For λ = 5,

A − 5I = ( \begin{matrix}−3 & 9 & 0 \\ 1 & −3 & −1 \\ 0 & 0 & 0 \end{matrix} ) ∼ ( \begin{matrix}1 & −3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} )

There is only one free variable in the echelon form system. So a basis for Nul(A − 5I) is

\{( \begin{matrix}3 \\ 1 \\ 0 \end{matrix} )\} .

Then, for λ = −1,

A + I = ( \begin{matrix}3 & 9 & 0 \\ 1 & 3 & −1 \\ 0 & 0 & 6 \end{matrix} ) ∼ ( \begin{matrix}1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} )

There is alson only one free variable in the echelon form system. So a basis for Nul(A + I) is

\{( \begin{matrix}−3 \\ 1 \\ 0 \end{matrix} )\} .

C) Is A diagonalizable? Explain.

Solution: A is not diagonalizable because

dim Nul(A − 5I) + dim Nul(A + I) = 2

(not 3).

VII. True - False. For each true statement give a brief proof. For each false statement give a counterexample or a reason. A) If y, z are two fixed vectors in R^{n}, then the mapping T : R^{n} → R^{n} defined by

T (x) = (x · y)y + (x · z)z

is linear.

Solution: This is True. To prove it, use the fact that the dot product is linear in the first variable.

B) No matter what the entries a, b, c ∈ R are, all the roots of the characteristic polynomial

of the matrix A = ( \begin{matrix}a & b \\ b & c \end{matrix} ) are real.

Solution: This is True. For the proof, use the quadratic formula to determine the roots of the characteristic polynomial. Both roots are real because the discriminant (the expression in the square root) is always ≥ 0. This fact was also derived in class – see the notes. (Or, you could also say: This follows from the Spectral Theorem since A is a symmetric matrix.)

C) If A ∈ M_{2×5}(R), then for every b ∈ R^{2} the general solution of the equation Ax = b contains three arbitrary constants.

Solution: This is False. The system could be inconsistent (so there are no solutions at all). For instance to make a counterexample, let

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

and let b be any vector in R^{2} whose second component is nonzero. Then b is not in the column space of A, and there are no solutions of Ax = b.

D) If A is an n × n matrix and det(A^{2} − 49I) = 0, then there exists a nonzero vector x such that Ax = 7x.

Solution: This is False. This would say that λ = 7 must be an eigenvalue of A. But here is a counterexample:

A = ( \begin{matrix}−7 & 0 \\ 0 & −7 \end{matrix} ) .

Note that λ = +7 is not an eigenvalue of A, but λ = 49 is an eigenvalue of

A^{2} = ( \begin{matrix}49 & 0 \\ 0 & 49 \end{matrix} ) .

E) If A, B ∈ M_{n×n}(R), and B = Q^{−1}AQ for some invertible n × n matrix Q, then det(B) = det(A).

Solution: This is True. For the proof, use standard properties of determinants (the product formula, the inverse formula).

F) If A ∈ M_{6×9}(R) then dim Col(A) can equal 8.

Solution: This is False. The column space of A is a subspace of R^{6}. No subspace of R^{6} has dimension larger than 6.

G) If A ∈ M_{6×9}(R) then dim Nul(A) is at most 6.

Solution: This is also False. Counterexample: If A is the 6 × 9 zero matrix, then dim Nul(A) = 9 (it’s all of R^{9}).

H) If A ∈ M_{6×9}(R) and dim Nul(A) = 4, then dim Col(A) = 5.

Solution: This is True. Proof uses the fact that if A is m × n, then

dim Nul(A) + dim Col(A) = n.

I) If U is a 7 × 3 matrix whose columns {u_{1}, u_{2}, u_{3}} form an orthonormal set in R^{7}, then u_{1}, u_{2}, u_{3} are eigenvectors of the 7 × 7 matrix UU^{t}.

Solution: This is True. They are all eigenvectors with eigenvalue 1. Because of the orthonormality, U^{t}u_{1} = e_{1}, U^{t}u_{2} = e_{2} and U^{t}u_{3} = e_{3} (the standard basis vectors in R^{3}. Therefore, UU^{t}u_{i} = Ue_{i} = u_{i} for i = 1, 2, 3.

J) If A is a 3 × 3 matrix with eigenvalues λ = 2, 5, −7 then Cramer’s Rule can be used to solve Ax = b for all b ∈ R^{3}.

Solution: This is True. The determinant of A is the value of the characteristic polynomial at λ = 0. Since we know the roots, the polynomial factors as

det(A − λI) = (2 − λ)(5 − λ)(−7 − λ).

Therefore det(A) = −70 ≠ 0. It follows that A is invertible, so Cramer’s Rule can be used to solve Ax = b no matter what b is.

VIII. (more challenging)

A) Show that there exists a basis {p_{1}(x), p_{2}(x), p_{3}(x)} for V = P_{2}(R) satisfying

\begin{matrix}p_{1}(0) = 1, & p_{1}(1) = 0, & p_{1}(2) = 0 \\ p_{2}(0) = 0, & p_{2}(1) = 1, & p_{2}(2) = 0 \\ p_{3}(0) = 0, & p_{3}(1) = 0, & p_{3}(2) = 1 \end{matrix}

Be sure to say how you know your polynomials form a basis for V.

Solution: To get a quadratic polynomial p_{1} that satisfies p_{1}(1) = p_{1}(2) = 0, note that x = 1, 2 are the roots. So the polynomial must factor as

p_{1}(x) = c(x − 1)(x − 2)

for some constant c. The constant must be chosen to get p_{1}(0) = 1, so c = 1/2. Hence p_{1}(x) can be written as

p_{1}(x) = \frac{(x − 1)(x − 2)}{(0 − 1)(0 − 2)}

Similarly, the other two polynomials are

p_{2}(x) = \frac{(x − 0)(x − 2)}{(1 − 0)(1 − 2)}

and

p_{3}(x) = \frac{(x − 0)(x − 1)}{(2 − 0)(2 − 1)}

To see that these are a basis for P_{2}(R), note that this vector space has dimension 3 (the standard basis {1, x, x^{2}} has three elements). Hence to show a set of three polynomials in here is a basis, we only have to show that it is linearly independent. If

0 = c_{1}p_{1}(x) + c_{2}p_{2}(x) + c_{3}p_{3}(x)

for all x, then substituting x = 0 gives

0 = c_{1} · 1 + c_{2} · 0 + c_{3} · 0.

Hence c_{1} = 0. Similarly, setting x = 1 and x = 2 shows c_{2} = c_{3} = 0 also.

B) Suppose you knew that the polynomial q(x) satisfied q(0) = 8, q(1) = −3 and q(2) = 4. What would the expansion of q(x) in terms of the basis {p_{1}(x), p_{2}(x), p_{3}(x)} be?

Solution: The expansion is

q(x) = 8p_{1}(x) + (−3)p_{2}(x) + 4p_{3}(x).

(Note what happens if you substitute x = 0, 1, 2 here.)

C) Generalize what you did in part A to show that given any three distinct real numbers a, b, c, there exists a basis {p_{1}(x), p_{2}(x), p_{3}(x)} of V = P_{2}(R) satisfying

\begin{matrix}p_{1}(a) = 1, & p_{1}(b) = 0, & p_{1}(c) = 0 \\ p_{2}(a) = 0, & p_{2}(b) = 1, & p_{2}(c) = 0 \\ p_{3}(a) = 0, & p_{3}(b) = 0, & p_{3}(c) = 1 \end{matrix}

Solution: Following the recipe from part A, the desired basis is

\{\frac{(x − b)(x − c)}{(a − b)(a − c)}, \frac{(x − a)(x − c)}{(b − a)(b − c)}, \frac{(x − a)(x − b)}{(c − a)(c − b)}\}

Note that a, b, c must be distinct or else we would be dividing by zero, and who knows what would happen then? The paper might burst into flames or something similarly dire! The given equations imply linear independence as in part A. Added Note: The p_{i}(x) are called Lagrange interpolating polynomials. If you are taking Numerical Analysis next year, you will use this idea a lot(!)

IX. (more challenging) A) Show that if x is an eigenvector of a matrix A with eigenvalue λ, then x is also an

eigenvector of A^{2}.

Solution: We have

A^{2}x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = λ^{2}x.

Hence x is an eigenvector of A with eigenvalue λ^{2}. B) Suppose that A is symmetric with eigenvalues λ_{1}, …, λ_{n}, and corresponding eigenvec-

tors x_{1}, …, x_{n} forming an orthonormal basis of R^{n}. Let P_{1}, …, P_{n} be the matrices of the orthogonal projections onto the lines spanned by the eigenvectors of A. Explain why

A^{2} = λ_{1}^{2}P_{1} + ⋯ + λ_{n}^{2}P_{n},

using the Spectral Theorem.

Solution: Using the fact that A is symmetric, the matrix A^{2} satisfies

(A^{2})^{t} = (AA)^{t} = A^{t}A^{t} = AA = A^{2}.

Therefore, A^{2} is also symmetric. The Spectral Theorem applied to A says that there is an orthonormal basis {u_{1}, …, u_{n}} of R^{n} consisting of eigenvectors of A. But part A shows that the u_{i} are also eigenvectors of A^{2}:

Au_{i} = λ_{i}u_{i} ⇒ A^{2}u_{i} = λ_{i}^{2}u_{i}.

The second part of the Spectral Theorem (the spectral decomposition) then shows

A^{2} = λ_{1}^{2}u_{1}u_{1}^{t} + ⋯ + λ_{n}^{2}u_{n}u_{n}^{t}.

The matrix P_{i} = u_{i}u_{i}^{t}, so this is what we wanted to show.

C) Verify that the formula in part B is true in the case of the 2 × 2 symmetric matrix

A = ( \begin{matrix}4 & 1 \\ 1 & 4 \end{matrix} ).

Solution: The spectral decomposition of A is

A = 5 · ( \begin{matrix}1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix} ) + 3 · ( \begin{matrix}1/2 & −1/2 \\ −1/2 & 1/2 \end{matrix} )

Then

5^{2} · ( \begin{matrix}1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix} ) + 3^{2} · ( \begin{matrix}1/2 & −1/2 \\ −1/2 & 1/2 \end{matrix} ) = ( \begin{matrix}17 & 8 \\ 8 & 17 \end{matrix} ) = A^{2}

which is what we wanted to show.