Cfrgtkky

$M$ $=$ $begin{pmatrix}3&2&2\ 2&3&2\ 2&2&3end{pmatrix}$. Diagonalize $M$ using an orthogonal matrix.

So I got that the eigenvalues for $M$ were $1$ and $7$. For the eigenvalue of $1$, I got the eigenvectors $begin{pmatrix}-1\ 0\ 1end{pmatrix}$ and $begin{pmatrix}-1\ 1\ 0end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $begin{pmatrix}1\ 1\ 1end{pmatrix}$. This gave me the diagonal matrix $begin{pmatrix}1&0&0\ 0&1&0\ 0&0&7end{pmatrix}$ and the orthogonal matrix $begin{pmatrix}-frac{1}{sqrt{2}}&-frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ 0&frac{1}{sqrt{2}}&frac{1}{sqrt{3}}\ frac{1}{sqrt{2}}&0&frac{1}{sqrt{3}}end{pmatrix}$.

But when I multiply the orthogonal matrix by the diagonal matrix and then its transpose, I get an answer that is slightly off what $M$ is, but I am not sure why.

If anyone knows where I may have gone wrong, I would greatly appreciate you telling me!

$$(-1, 0, 1) cdot (-1, 1, 0)=1$$

They are not orthogonal.

Just do a gram-schmidt step to find a set of orthogonal eigenvectors for eigenvalues corresponding to $1$.

start with

$$
left(
begin{array}{rrr}
1 & -1 & -1 \
1 & 1 & -1 \
1 & 0 & 2 \
end{array}
right).
$$
and divide the columns by $sqrt 3, sqrt 2, sqrt 6$

If you had the analogous problem in 4 by 4, you could begin with

$$
left(
begin{array}{rrrr}
1 & -1 & -1 & -1 \
1 & 1 & -1 & -1 \
1 & 0 & 2 & -1 \
1 & 0 & 0 & 3 \
end{array}
right).
$$
and divide the columns by $2,sqrt 2, sqrt 6, sqrt {12}$

for 5 by 5
$$
left(
begin{array}{rrrrr}
1 & -1 & -1 & -1 & -1 \
1 & 1 & -1 & -1 & -1 \
1 & 0 & 2 & -1 & -1 \
1 & 0 & 0 & 3 & -1 \
1 & 0 & 0 & 0 & 4 \
end{array}
right).
$$
$sqrt 5,sqrt 2, sqrt 6, sqrt {12}, sqrt{20}$