A First Course in Linear Algebra, 2017a

571
7.1.2 Say AX = λX.ThencAX = cλX and so the eigenvalues of cA are just cλ where λ is an eigenvalue
of A.
7.1.3 BAX = ABX = AλX = λAX. Here it is assumed that BX = λX.
7.1.4 Let X be the eigenvector. Then A m X = λ m X,A m X = AX = λX and so
λ m = λ
Hence if λ ≠ 0, then
and so |λ| = 1.
λ m−1 = 1
7.1.5 The formula follows from properties of matrix multiplications. However, this vector might not be
an eigenvector because it might equal 0 and eigenvectors cannot equal 0.
7.1.14 Yes.
[
0 1
0 0
]
works.
7.1.16 When you think of this geometrically, it is clear that the only two values of θ are 0 and π or these
added to integer multiples of 2π
7.1.17 The matrix of T is
7.1.18 The matrix of T is
7.1.19 The matrix of T is ⎣
[ 1 0
0 −1
[ 0 −1
1 0
⎡
]
. The eigenvectors and eigenvalues are:
{[
0
1
1 0 0
0 1 0
0 0 −1
⎧⎡
⎨
⎣
⎩
]}
{[
1
↔−1,
0
]}
↔ 1
]
. The eigenvectors and eigenvalues are:
{[ −i
1
0
0
1
⎤
]}
{[ i
↔−i,
1
]}
↔ i
⎦ The eigenvectors and eigenvalues are:
⎤⎫
⎧⎡
⎬ ⎨
⎦
⎭ ↔−1, ⎣
⎩
1
0
0
⎤
⎡
⎦, ⎣
0
1
0
⎤⎫
⎬
⎦
⎭ ↔ 1
7.2.1 The eigenvalues are −1,−1,1. The eigenvectors corresponding to the eigenvalues are:
⎧⎡
⎤⎫
⎧⎡
⎤⎫
⎨ 10 ⎬ ⎨ 7 ⎬
⎣ −2 ⎦
⎩ ⎭ ↔−1, ⎣ −2 ⎦
⎩ ⎭ ↔ 1
3
2
Therefore this matrix is not diagonalizable.