Mathematics 244 – Linear Algebra Answers – Final Exam May 11, 2004

I.

A) Let p, q ∈ W. Then p′(1) = q′(1) = 0, p(0) = 3p(2), and q(0) = 3q(2). So if c is any constant, then (cp + q)′(1) = cp′(1) + q′(1) = c · 0 + 0. Also,

(cp + q)(0) = cp(0) + q(0) = c · 3p(2) + 3q(2) = 3(cp(2) + q(2)) = 3(cp + q)(2).

Since W is closed under sums and scalar multiples, W is a subspace of P_{4}(R).

B) Let p = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} be a general element of P_{4}(R). Then p ∈ W if and only if

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

This says that the coefficients of p must satisfy

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

Putting the system into echelon form, we get

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

The free variables are a_{2}, a_{3}, a_{4}, so by the usual process we get

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

(the first comes by setting a_{2} = 1, a_{3} = 0, a_{4} = 0, etc.)

C) This set is linearly independent, since if c_{1}(x^{2}−2x)+c_{2}(x^{3}−3x−3)+c_{3}(x^{4}−4x−12) = 0, then equating coefficients of x^{2} shows c_{1} = 0, equating coefficients of x^{3} shows c_{2} = 0, and equating coefficients of x^{4} shows c_{3} = 0.

II.

See class notes. The proof of (1.6.8) in the text follows almost the same reasoning as this proof.

III. A)

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

since T (e_{1}) = (4, 0, 1) = 4e_{1} + 0e_{2} + 1e_{3}, etc.

B) We have

[T]_{α′}^{β′} = [I_{R^{3}}]_{std}^{β′}[T]_{std}^{std}[I_{R^{4}}]_{α′}^{std}

From the given bases α′, β′,

[I_{R^{4}}]_{α′}^{std} = ( \begin{matrix}1 & 1 & 0 & 0 \\ −1 & 1 & 0 & 0 \\ 0 & 0 & 2 & −1 \\ 0 & 0 & 3 & −2 \end{matrix} )

By the Gauss-Jordan procedure,

[I_{R^{3}}]_{std}^{β′} = ( \begin{matrix}0 & 0 & −1 \\ 0 & −1 & 1 \\ 1 & 1 & 0 \end{matrix} )^{−1} = ( \begin{matrix}1 & 1 & 1 \\ −1 & −1 & 0 \\ −1 & 0 & 0 \end{matrix} )

So,

[T]_{α′}^{β′} = ( \begin{matrix}1 & 1 & 1 \\ −1 & −1 & 0 \\ −1 & 0 & 0 \end{matrix} ) ( \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}7 & 3 & 0 & −7 \\ −6 & −2 & −3 & 5 \\ −7 & −1 & −1 & −4 \end{matrix} )

IV. The corresponding echelon form system is

( \begin{matrix}1 & 0 & 6/7 & −1/7 & | & −2/7 \\ 0 & 1 & 3/7 & 3/7 & | & 6/7 \\ 0 & 0 & 0 & 0 & | & a − 10 \end{matrix} )

This means the system is consistent if and only if a = 10. If so the solutions of the system are all vectors

(\frac{−2}{7}, \frac{6}{7}, 0, 0) + s (\frac{−6}{7}, \frac{−3}{7}, 1, 0) + t (\frac{1}{7}, \frac{−3}{7}, 0, 1)

where s, t are arbitrary real numbers.

V. The base case is when A = (a_{11}). By definition, det(A) = a_{11} is the product of the diagonal entries in this case. Now, assume that the result is true for all k × k uppertriangular matrices and consider an upper-triangular A which is (k+1)×(k+1). expanding along the first column, we see det(A) = a_{11}det(A_{11})+0, since all the other entries in column 1 are zeroes. The (1,1)-minor A_{11} is a k × k upper-triangular matrix with diagonal entries a_{22}, …, a_{k+1,k+1}, so the induction hypothesis shows that

det(A) = a_{11} det(A_{11}) = a_{11}a_{22} ⋯ a_{k+1,k+1}

which establishes the induction step. Hence the result is true for all upper-triangular matrices.

VI. A) Expanding det(A − λI) along row 3, we have

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

so the eigenvalues are λ = 5, 5, −1.

B) We have

E_{5} = Ker ( \begin{matrix}−3 & 9 & 0 \\ 1 & −3 & −1 \\ 0 & 0 & 0 \end{matrix} )

= Ker ( \begin{matrix}1 & −3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} )

= Span{(3, 1, 0)}

Similarly,

E_{−1} = Ker ( \begin{matrix}3 & 9 & 0 \\ 1 & 3 & −1 \\ 0 & 0 & 0 \end{matrix} )

= Ker ( \begin{matrix}1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} )

= Span{(−3, 1, 0)}

C) The multiplicity of λ = 5 as a root of the characteristic polynomial is 2, but dim E_{5} = 1. Hence there is no way to get a linearly independent set of 3 eigenvectors of this A. Hence A is not diagonalizable.

VII. A) True: Because of the bilinearity of the dot product,

T (cx + w) = ⟨cx + w, y⟩y + ⟨cx + w, z⟩z
= c⟨x, y⟩y + ⟨w, y⟩y + c⟨x, z⟩z + ⟨w, z⟩z
= c(⟨x, y⟩y + ⟨x, z⟩z) + ⟨w, y⟩y + ⟨w, z⟩z
= cT (x) + T (w)

B) True, since the matrix is symmetric (see (4.5.6) in the text). (Many people worked much, much too hard on this one!)

C) False. If T is not surjective, then dim Im(T) < 2 and the Dimension Theorem says dim Ker(T) > 3. In this case, then

\{ \begin{matrix}y ∉ Im(T) ⇒ & no solutions at all \\ y ∈ Im(T) ⇒ & solutions contain more than 3 arbitrary consts. \end{matrix}

For example, say T (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (x_{1}, x_{1}) is a mapping T : R^{5} → R^{2}. The image consists of all vectors whose components are equal. If (1, 1) is such a vector then T (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (1, 1) has general solution

(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (1, 0, 0, 0, 0)+s(0, 1, 0, 0, 0)+t(0, 0, 1, 0, 0)+u(0, 0, 0, 1, 0)+v(0, 0, 0, 0, 1)

where s, t, u, v are arbitrary. For other vectors like (2, 3), T (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (2, 3) has no solutions at all.

VIII. A) The main point of this problem is to derive a way to write down the desired polynomials. The most economical way to do this is to note that p_{1} must have roots at x = 1, 2, p_{2} must have roots at x = 0, 2, etc. This says

p_{1}(x) = c_{1}(x − 1)(x − 2), p_{2}(x) = c_{2}x(x − 2), p_{3}(x) = c_{3}x(x − 1)

for some constants c_{i}. Then the values of c_{i} can be determined by substituting the other x-value and using the given information

p_{1}(0) = 1 ⇒ c_{1}(0 − 1)(0 − 2) = 1 ⇒ c_{1} = 1/2

and similarly, c_{2} = −1, c_{3} = 1/2. The polynomials are

\begin{aligned}p_{1}(x) & = \frac{1}{2}x^{2} − \frac{3}{2}x + 1 \\ p_{2}(x) & = −x^{2} + 2x \\ p_{3}(x) & = \frac{1}{2}x^{2} − \frac{1}{2}x \end{aligned}

To check that these are linearly independent, we consider a linear combination,

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

Substitute x = 0, then we get 0 = a_{1} ·1+a_{2} ·0+a_{3} ·0. Hence a_{1} = 0. Similarly, substituting x = 1 shows a_{2} = 0, x = 2 shows a_{3} = 0. Since dim P_{2}(R) = 3, and this is an independent subset with 3 elements, it is a basis.

B) Write

q(x) = a_{1}p_{1}(x) + a_{2}p_{2}(x) + a_{3}p_{3}(x)

As above, substitute x = 0, 1, 2 one after another. x = 0 gives

8 = q(0) = a_{1} · 1 + a_{2} · 0 + a_{3} · 0

Hence a_{1} = 8. Similarly, a_{2} = −3, a_{3} = 4.

C) Following the idea of proof given in part A above,

p_{1}(x) = \frac{(x − b)(x − c)}{(a − b)(a − c)}

p_{2}(x) = \frac{(x − a)(x − c)}{(b − a)(b − c)}

p_{1}(x) = \frac{(x − a)(x − b)}{(c − a)(c − b)}

This is a special case of a general result called the Lagrange Interpolation Theorem.

VIII’. A) If Ax = λx, then

A^{2}x = A(Ax) = A(λx) = λAx = λ^{2}x

Hence x is also an eigenvector of A^{2} with eigenvalue λ^{2}.

B) Since A is symmetric, so is A^{2} (reason: (A^{2})^{t} = (AA)^{t} = A^{t}A^{t} = A^{2}). Apply the Spectral Theorem to the symmetric matrix A^{2}. Part A says that the eigenspaces and the projections for A^{2} are the same as those for A (the eigenspaces are equal). But the eigenvalues for A^{2} are the λ^{2}. Hence the spectral decomposition for A^{2} is

A^{2} = λ_{1}^{2}P_{1} + ⋯ + λ_{k}^{2}P_{k}

(Alternate method: From the spectral decomposition for A,

A = λ_{1}P_{1} + ⋯ + λ_{k}P_{k}

So

A^{2} = (λ_{1}P_{1} + ⋯ + λ_{k}P_{k})(λ_{1}P_{1} + ⋯ + λ_{k}P_{k})

Matrix product is distributive over matrix sum, so this can be written as

A^{2} = λ_{1}^{2}P_{1}^{2} + ⋯ + λ_{k}^{2}P_{k}^{2} + ∑_{i≠j} λ_{i}λ_{j}P_{i}P_{j}

But P_{i}^{2} = P_{i} and if i ≠ j, then P_{i}P_{j} = 0. This follows because as we know, eigenvectors for distinct eigenvalues of a symmetric matrix are orthogonal, hence Im(P_{j}) = E_{λ_{j}} ⊂ (E_{λ_{i}})^{⊥} = Ker(P_{i}). So when you apply P_{i}P_{j} to any vector you will get zero.)

C) The eigenvalues of A are λ = 3, 5. Correxponding bases for the eigenspaces are {(1, 1)} for E_{5} and {(−1, 1)} for E_{3}.

P_{E_{5}} = ( \begin{matrix}1 \\ 1 \end{matrix} ) (( 1 1 ) ( \begin{matrix}1 \\ 1 \end{matrix} ))^{−1} ( 1 1 ) = ( \begin{matrix}1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix} )

Similarly,

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

So

A^{2} = ( \begin{matrix}17 & 8 \\ 8 & 17 \end{matrix} ) = 25 ( \begin{matrix}1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix} ) + 9 ( \begin{matrix}1/2 & −1/2 \\ −1/2 & 1/2 \end{matrix} )