Question

Let A is a 4x4 matrix such that
A(0,1,0,1)^t=(0,1,0,1)^t and
A(1,0,1,0)^t=(-1,0,-1,0)^t. Also, suppose A has trace 4 and det 9/4.
Find all eigenvalues of A and their multiplicites and explain why (1,1,1,1) is not an eigenvector of A.

Answer 1

the two equations
A(0,1,0,1)^t=(0,1,0,1)^t and
A(1,0,1,0)^t=(-1,0,-1,0)^t.

tell you 2 of the eigenvalues (and vectors): 1 and -1
the trace is the sum of the eigenvalues
the set is the product of the eigenvalues

Answer 2

**the determinant is the product of the eigenvalues

Answer 3

All i'm confused about it the information at the beginning

Answer 4

of the question

Answer 5

how do you already know two eigen values?

Answer 6

A(0,1,0,1)^t=(0,1,0,1)^t
where t means transpose
is a way to write
\[A \left[\begin{matrix}0 \\ 1\\0 \\1\end{matrix}\right]= 1 \cdot \left[\begin{matrix}0 \\ 1\\0 \\1\end{matrix}\right]\]

Answer 7

yea i understood that before.

Answer 8

which should look familiar:
\[ A v = \lambda v\]

Answer 9

ohhhhhh

Answer 10

yep. so the other would be -1. get ya

Answer 11

yep now its easy to get lamda3 and lamda4

Answer 12

yes. for the last part, [1 1 1 1] is a linear combination of the two eigenvectors they give you.

Answer 13

yea. it looked obvious by looking at it from the beginning. thanks for starting me off!

Answer 14

alright. so i found that lamda=1,-1,9/2,-1/2

Answer 15

then does this mean their multiplicities are just 1?

Answer 16

yes, all eigenvalues are distinct, so you have 4 eigenvalues, with multiplicity of 1 for each
(a repeated value would give a multiplicity > 1)

Answer 17

yea. how would you represent in mathematical terms? or is what you wrote sufficient?

Answer 18

what I wrote should be sufficient

Answer 19

so with b) (1,1,1,1)=(0,1,0,1)+(1,0,1,0) hence this is a linear combination of the two basis hence not an eigen vector?

Answer 20

yes

Answer 21

is it right to say that (0,1,0,1) and (1,0,1,0) are eigenbasis's?

Answer 22

yes

Answer 23

but each of them are eigen vectors?

Answer 24

yes, the 4 distinct eigenvectors (because you have 4 distinct eigenvalues) make up a basis in 4d

Answer 25

legend.

Answer 26

thanks phi

Answer 27

thanks phi