HW 13: Some problems
5.1
• #25.
Proof. Suppose λ is an eigenvalue of A. Since A is invertible, we must
have λ 6= 0. We then can find ~x 6= ~0 so that
A~x = λ~x
Multiplying A−1 (which exists since A is invertible) on both sides, we thus
have:
A−1 A~x = A−1 (λ~x) = λA−1 ~x
which tells us that:
A−1 ~x = λ−1 ~x
Since ~x 6= 0, λ−1 is one eigenvalue of A−1 .
Remark: Some people did like this: (A~x)−1 = (λ~x)−1 = λ−1 ~x−1 or
~x/(λ~x) = A−1 (λ~x)/(λ~x). These are totally wrong! How can you divide by
a vector?
• #26
Proof. Suppose λ is an e-val of the matrix. Then, we find ~x 6= ~0, so that
A~x = λ~x
Multiplying this by A again, we can have:
~0 = A2 ~x = λ2 ~x
Therefore, λ2 = 0 and λ = 0.
Remark: Using the same method, you can show that if An = In , the only
reals roots of A could be ±1. Some people said that if A2 = 0, the entries
on the diagonal must be 0. This is wrong! See for example:


1
1
2
1
2 
A= 1
−1 −1 −2
1
• #31
We need the vector reflected, namely A~v to be in the same line with ~v
(since A~v = λ~v ). We can see that there are only two possibilities for this
requirement:
Case 1: If ~v is on the axis of reflection, then A~v = ~v . therefore, λ = 1 and
the eigenspace is the axis of reflection in this case.
Case 2: If ~v is perpendicular to the axis of reflection, then A~v = −~v .
Therefore, λ = −1 and the eigen-space is the line perpendicular to the
axis and going through origin in this case.
5.2
# 20.
Proof. Since (A − λ)T = AT − λ for any real value λ. Then, we have:
det(A − λ) = det((A − λ)T ) = det(AT − λ)
This exactly is saying that A and AT have the same characteristic polynomial.
Extra 2
(a). the characteristic polynomial is easy to compute
det(A − λ) = λ2 − 2 cos θλ + 1
(b). The discriminant b2 − 4ac must be nonnegative to have real roots.
Therefore, (−2 cos θ)2 − 4 ≥ 0. This implies sin θ = 0. Therefore, θ = 2kπ or
θ = 2kπ + π.
(c). When θ = 2kπ, the matrix is I2 , which is already in diagonal form and
thus it’s diagonalizable. (Let D = A, P = I2 ). When, θ = 2kπ + π, A = −I2 .
Also in diagonal form and thus diagonalizable. (Let D = A, P = I2 ).
Remark: Some people just lost the case that cos θ could −1.
Some extra
a. Write out a 2 × 2 matrix that is not diagonalizable.
b. Write out a 4 × 4 matrix so that it has 2 distinct real eigenvalues but no
complex eigenvalues so that it’s not diagonalizable.
2