Is congruent diagonalization of symmetric matrix necessarily a similar one?

Question

It can be shown that for real symmetric matrix $A$, there is an orthogonal matrix $Q$ such that $$Q^TAQ=Q^{-1}AQ=\Lambda,$$ where $\Lambda$ is the diagonal matrix whose entries are $A$'s eigenvalues.

My question arises while considering the "converse": if $Q^TAQ=\Lambda$ for real symmetric $A$, and $\Lambda$ again denotes the diagonal matrix of eigenvalues, can one claim that $Q$ is orthogonal?

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I think it's not true at the first glance, since there are many choices that we can congruently transform $A$ to a diagonal matrix. But things may get subtle after confining $\Lambda$ to be a special matrix consisting of the eigenvalues of $A$.

I can derive $(QC)^TA(QC)=C^T\Lambda C$ for some matrix $C$, maybe we can choose some $C$ to yield a counterexample? Or we can actually prove it? Thank you!

Answer 1

For instance, $$\begin{pmatrix}0& \sqrt{\frac12}\\ \sqrt2&0\end{pmatrix}\begin{pmatrix}1&0\\0 &2\end{pmatrix}\begin{pmatrix}0& \sqrt2\\ \sqrt{\frac12}&0\end{pmatrix}=\begin{pmatrix}1& 0 \\ 0& 2\end{pmatrix}$$

Answer 2

Here is a generalisation of Saucy's counterexample. Pick any orthogonal matrix $V$ with some nonzero off-diagonal entry $v_{ij}$. Pick any positive diagonal matrix $\Lambda$ such that $|v_{ij}\sqrt{\lambda_j/\lambda_i}|>1$. Then $Q=\Lambda^{-1/2}V\Lambda^{1/2}$ is not an orthogonal matrix (because $|q_{ij}|=|v_{ij}\sqrt{\lambda_j/\lambda_i}|>1$) but $Q^T\Lambda Q=\Lambda$.