Why diagonal matrix doesn't change eigenvalues?

Printable View

August 18th 2011, 07:41 PM
xEnOn

Why diagonal matrix doesn't change eigenvalues?

Suppose I have this: $A=P_{0}\Lambda P_{0}^{-1}$
And then there is a diagonal matrix $D$ such that its determinant is always equals to 1: $det(D)=1$

Then, for some reason, $DA = P \Lambda P^{-1}$ is always true.

How can I show that for the lambda, which is the eigenvalues matrix, does not change for all $DA$ ?

And what is the relationship between $P$ and $P_{0}$ ? I understand that their determinant $det(P_{0}\Lambda P_{0}^{-1})=det(P \Lambda P^{-1})$ but still, what's the relationship between the $P_{0}$ and $P$ because $P_{0}$ is eigenvectors matrix for $A$ while $P$ is eigenvectors matrix for $DA$ and so they are different. But still, there isn't any strong relationship, is there?

Thanks! :)
August 19th 2011, 07:03 AM
HallsofIvy

Re: Why diagonal matrix doesn't change eigenvalues?

If I am understanding you correctly, this is not true.

For example, suppose $A= \begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}$ . It is easy to see that A has eigenvalues 2 and 3.

Now, take $D= \begin{bmatrix}\frac{3}{2} & 0 \\ 0 & 2\end{bmatrix}$ , a diagonal matrix with determinant 1.

$DA= \begin{bmatrix}-\frac{3}{2} & 2 \\ -12 & 12\end{bmatrix}$
which does NOT have 2 and 3 as eigenvalues.
August 20th 2011, 01:37 AM
xEnOn

Re: Why diagonal matrix doesn't change eigenvalues?

hmm...This is weird. Earlier on, I tried having several matrices having my $D=\begin{bmatrix}2 & 0\\ 0 & \frac{1}{2}\end{bmatrix}$ and then many different matrices for $A$ such as $A=\begin{bmatrix}1 & 3\\ 4 & 2\end{bmatrix}$ , $A=\begin{bmatrix}1 & 14\\ 8 & 2\end{bmatrix}$ , $A=\begin{bmatrix}1 & 2\\ 3 & 2\end{bmatrix}$ , etc etc, they all worked.

After your example, I tried more values into the matrices and began to realise that it doesn't work all the time. I'm curious if there is any kind of special properties that allow this proposition to be true? (Wondering)