For every matrix $A$ we have
$$AV=V\Lambda\rightarrow A=V\Lambda V^{-1}$$
where $V$ is the matrix of eigenvectors and $\Lambda$ is the diagonal matrix including eigenvalues.
When we have a symmetric matrix, one can write
$$A^T=({V^{-1}})^T\Lambda V^T\rightarrow A=({V^{-1}})^T\Lambda V^T$$
where symmetry of $A$ results in $A^T=A$, and $\Lambda^T=\Lambda$ because it is diagonal.
Comparing these two yields
$$V^{-1}=V^T\rightarrow V^TV=I$$
which is the definition of the orthogonal matrix. That means not only eigenvectors of a symmetric matrix are orthogonal but also they are orthonormal, i.e., they are unit vector with length 1 and orthogonal.
Is this correct for all symmetric matrices?
Best Answer
That step is wrong. I mean, it's not a bad guess to make. But suppose for a moment that you've found a $V$ such that $V^T = V^{-1}$.
Now write $W = 2V$. Then we'll have $$ A = W \Lambda W^{-1} $$ because the factors of $2$ and $\frac{1}{2}$ will cancel.
But we'll also have $$ A = (W^T)^{-1} \Lambda W^T $$ by the reasoning you presented. If the quoted remark above were valid, you would conclude that $W$ was ALSO orthonormal, which is impossible, because it's twice an orthonormal matrix.
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It's true, however, that by taking the matrix $V$ of eigenvectors, you can produce an orthonormal matrix of eigenvectors: you just divide each column by its norm.
So your conclusion, that a symmetric matrix can be written as $U^t \Lambda U$, where $U$'s columns are an orthonormal basis of eigenvectors, is correct, but your method of reaching that conclusion was flawed.