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$M$ $=$ $\begin{pmatrix}3&2&2\\ 2&3&2\\ 2&2&3\end{pmatrix}$. 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 $\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}$ and $\begin{pmatrix}-1\\ 1\\ 0\end{pmatrix}$, and for the eigenvalue of $7$, I got the eigenvector $\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}$. This gave me the diagonal matrix $\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&7\end{pmatrix}$ 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!

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    $\begingroup$ Both eigenvectors for $\;\lambda=1\;$ are wrong, as you can easily check. The eigenvectors corresponding to different eigenvalues are orthogonal (because the matrix is symmetric), so you must only do GM in each eigenspace... $\endgroup$ – DonAntonio Nov 19 '18 at 2:35
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$$(-1, 0, 1) \cdot (-1, 1, 0)=1$$

They are not orthogonal.

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

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

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