<s>[INST] Orthogonally diagonalizable matrix If $A$ a $3\times3$ matrix, where
$A \begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}=3\begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}$, $A \begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}=3\begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}$, $A \begin{bmatrix} -6 \\ 1\\ 1\\ \end{bmatrix}=2\begin{bmatrix} -6 \\ 1\\ 1\\ \end{bmatrix}$.
I need to check if $A$ is orthogonally diagonalizable. How would I do that by these relations?
 [/INST] Recall that a real square matrix is orthogonally diagonalizable if and only if $A$ is a symmetric matrix. (The forwards implication is easy, the reverse implication is the spectral theorem).
Note that $$A\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}=\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}\begin{bmatrix} 3&0&0\\0&3&0\\0&0&2 \end{bmatrix}.$$
From this you can calculate $A$ explicitly by multiplying with the inverse (on the right) of the matrix consisting of eigenvectors.
Alternatively, you can try to verify directly whether or not the eigenvectors are orthogonal to each other. If so, the jobs done (and this is clearly the case here).
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