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Let $A\in M_n(\mathbb R)$ be a symmetric matrix and $V=\mathbb R^n$. Let $T:\mathbb R^n \rightarrow \mathbb R^n$ be the operator such that $[T]_\alpha = A$, where $[T]_\alpha$ is the matrix of $T$ in the canonical basis $\alpha = \{e_1, \cdots, e_n\}$. If $f$ is the bilinear form such that $[f]_\alpha = A$, where $[f]_\alpha = [f(e_i,e_j)]_{i,j}$, prove or disprove:

If $\beta$ is another basis of $V$ such that $[f]_\beta$ is diagonal, then $[T]_\beta$ is also diagonal?

Since $\beta$ is an arbitrary basis, it seems to be false. If we have $[f]_\beta$ diagonal, then there exists $P$ such that $P^tAP = [f]_\beta$. If $P$ is orthogonal, then $P^t=P^{-1}$ implies that $[T]_\beta = P^{-1}AP = [f]_\beta $, so $[T]_\beta$ is diagonal. But the matrix $P$ is not necessarily orthogonal for any basis $\beta$. Can anyone provide a counterxample where this fails? That is, exhibit a matrix $A$ and a basis $\beta$ such that the form $[f]_\beta$ is diagonal, but $[T]_\beta$ is not diagonal?

user2345678
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  • A symmetric matrix is always orthogonally diagonalizable. – azif00 Mar 10 '21 at 23:18
  • @azif00 I know that, but here it is not given that $\beta$ is an orthogonal basis of eigenvectors of $T$. $\beta$ could be any basis – user2345678 Mar 10 '21 at 23:19
  • Oh, you're right. My mistake! Let me think... – azif00 Mar 10 '21 at 23:20
  • it would be prudent to work out a $2\times 2$ case where all eigenvalues are distinct and $A$ is not diagonal. The existence of some non-orthogonal mapping to $\beta$ (e.g. in $LDL^T$ factorization) that diagonalizes $A$ via congruence, should do it. Alternatively it could be helpful to work over $\mathbb C$ and notice that complex symmetric (not hermitian) matrices are always congruent to diagonal matrices yet can be defective -- complex symmetric may be awkward but algebraically the mechanics for changing basis wrt to similarity and congruence transforms are the same as in reals. – user8675309 Mar 11 '21 at 00:32

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