MATH 141: Linear Analysis I Homework 13 solved


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1. Find all eigenvectors and eigenvalues of the matrix B =

1 0 1
0 4 0
12 2 2

. Write 1-3 sentences that interpret
these geometrically—in other words, what do the eigenvectors and eigenvalues tell you about the transformation geometrically (in terms of stretch factors and stretch directions)?
2. Construct an example for each of the following, or explain why such an example does not exist:
(a) a 2×2 matrix that is invertible but not diagonalizable
(b) a 2×2 non-diagonal matrix that is diagonalizable but not invertible
3. (Strang §5.2 #11) If all eigenvalues of A are 1, 1, and 2, which of the following are certain to be true? Give a
reason if true or a counterexample if false.
(a) A is invertible.
(b) A is diagonalizable.
(c) A is not diagonalizable.
4. (Strang §5.2 #12) Suppose the only eigenvectors of A are multiples of ~x =


, which of the following are
certain to be true? Give a reason if true or a counterexample if false.
(a) A is not invertible.
(b) A has a repeated eigenvalue.
(c) A is not diagonalizable.
5. (Strang §5.1 #18) Suppose a 3×3 matrix A has eigenvalues 0, 3, and 5 with associated eigenvectors ~u, ~v, and
~w respectively.
(a) Since the eigenvalues of A are all distinct, the set {~u, ~v, ~w} is .
(b) Write down a basis for the nullspace N(A) and the column space C(A).
(c) Find one particular solution to A~x = ~v + ~w. Find all solutions to A~x = ~v + ~w.
(d) Explain why A~x = ~u does not have a solution. (Hint: If there is a solution, then is in C(A).
Explain why that is impossible.)
(e) Is A invertible? Why or why not?
6. Let A be an n-by-n matrix. Suppose A~u = 2~u and A~v = 5~v for nonzero vectors ~u and ~v. Complete the
following proof that {~u, ~v} is linearly independent.
Proof: In order for {~u, ~v} to be linearly independent, we need to show that
the only solution to the vector equation x~u + y~v = ~0 is the trivial one.
Note that in this equation, ~u and ~v are known vectors, while x and y are unknown scalars. Now assume
that x = a and y = b is some solution to the above equation. That is
a~u + b~v = ~0. (1)
MATH 141: Linear Analysis I Homework 13
Multiply both sides of equation (1) by matrix A from the left. Show your calculation details to explain why
the following has to be true as well
2a~u + 5b~v = ~0. (2)
Explain in detail how combing vectors equations (1) and (2) leads to the conclusion that a = 0 = b. Therefore, the only solution to x~u + y~v = ~0 is the trivial solution.
7. In one of the reflection questions, you saw that if A is 2×2 matrix then the product of its eigenvalues is
equal to det(A) and the sum is equal to trace(A). The following shows that both claims still hold true for
n×n matrices.
Suppose that λ1, λ2, . . . , λn are the n eigenvalues of an n×n matrix A. λi
’s are the roots of the polynomial
det(A − λI), which means that we have a factorization
det(A − λI) = (λ1 − λ)(λ2 − λ)· · ·(λn − λ) (3)
(a) (Strang §5.1 #8) By making a clever choice of the value for λ in equation (3), show that det(A) is equal
to the product of eigenvalues.
(b) (Strang §5.1 #9) Show that trace(A) is equal to the sum of eigenvalues in three steps. First, find the
coefficient of (−λ)
n−1 on the righthand side of equation (3). Next, find all terms on the righthand side
of the following that involves (−λ)
det(A − λI) = det

a11 − λ a12 · · · a1n
a21 a22 − λ · · · a2n
an1 an2 · · · ann − λ

where aij are the entries of A. Add up all those terms to find the coefficient of (−λ)
. Lastly,
compare the coefficient of (−λ)
found in these two different ways.