Question

If A is a diagonizable matrix, how do the eigenvalues of A compare to the eigenvalues of A^2?

Answer 1

square them

Answer 2

are you sure they dont just remain the same? because they'll have the same charactersitic poloynomial?

Answer 3

simple example of why they should change....

let \[A=\left[\begin{matrix}2& 0 \\ 0& 3\end{matrix}\right]\]

\[A^2=\left[\begin{matrix}4& 0 \\ 0& 9\end{matrix}\right]\]

what are the eigenvalues of \(A\) and \(A^2\)

Answer 4

ok right, thanks! can i also ask what the comparison would be of the eigen values of A and the eigen values of A inverse

Answer 5

if \[A=PDP^{-1}\]

then \[A^{-1}=(PDP^{-1})^{-1}=(P^{-1})^{-1}D^{-1}P^{-1}=PD^{-1}P^{-1}\]

Answer 6

if \[D=\left[\begin{matrix}d_1 &0&0\\0&\ddots & 0 \\0&0&d_n\end{matrix}\right]\]

\[D^{-1}=\left[\begin{matrix}\frac{1}{d_1} &0&0\\0&\ddots & 0 \\0&0&\frac{1}{d_n}\end{matrix}\right]\]

Answer 7

thank you sooooo much!!!!!

Answer 8

yw

Answer 9

wait sorry can i ask one mpore question, so the eigen values of the inverse of A would be just the inverse of the eignen values from A?

Answer 10

the eigenvalues of A inverse are the reciprocals of the eigenvalues of A

Answer 11

right, ok, awesome!