1. ## Does matrix AS=SD?

This is part of a bigger question, though I've answered much of the parts before this part. I'll list the data before I list the question.

$A=\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 & 0\end{bmatrix}$
$S=\begin{bmatrix}1 & 1 & 0\\1 & 0 & 1\\1 & -1 & -1\end{bmatrix}$
(The columns of $S$ are the eigenvectors of $A$, just to note)

Now onto the question.

Let $D$ be the diagonal matrix whose entry $d_i$ is the eigenvalue corresponding to column $i$ of $S$. Compute $SD$. Does $AS=SD$?

It's probably not that hard, but I'm a little short on time due to other matters. I'm also not sure whether or not there was a typo on the question we were given.

2. Originally Posted by Runty
This is part of a bigger question, though I've answered much of the parts before this part. I'll list the data before I list the question.

$A=\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 & 0\end{bmatrix}$
$S=\begin{bmatrix}1 & 1 & 0\\1 & 0 & 1\\1 & -1 & -1\end{bmatrix}$
(The columns of $S$ are the eigenvectors of $A$, just to note)

Now onto the question.

Let $D$ be the diagonal matrix whose entry $d_i$ is the eigenvalue corresponding to column $i$ of $S$. Compute $SD$. Does $AS=SD$?

It's probably not that hard, but I'm a little short on time due to other matters. I'm also not sure whether or not there was a typo on the question we were given.

Well, as there are three lin. ind. eigenvectors of A (this follows from calculating the determinant of S), we know that $S^{-1}AS=D$ ...and from here you can solve the question .

Tonio