# Orthogonal Diagonalization of a $3$ by $3$ Matrix

$$M$$ $$=$$ $$\begin{pmatrix}3&2&2\\ 2&3&2\\ 2&2&3\end{pmatrix}$$. 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!

• 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... – DonAntonio Nov 19 '18 at 2:35

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

They are not orthogonal.

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

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