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The thought came from the following problem:

Let $V$ be a Euclidean space. Let $T$ be an inner product on $V$. Let $f$ be a linear transformation $f:V \to V$ such that $T(x,f(y))=T(f(x),y)$ for $x,y\in V$. Let $v_1,\dots,v_n$ be an orthonormal basis, and let $A=(a_{ij})$ be the matrix of $f$ with respect to this basis.

The goal here is to prove that the $A$ is symmetric. I can prove this easily enough by saying:

Since $T$ is an inner product, $T(v_i,v_j)=\delta_{ij}$.

\begin{align*} T(A v_j,v_i)&=T(\sum_{k=1}^n a_{kj} v_k,v_i)\\ &=T(a_{1j} v_1,v_i) + \dots + T(a_{nj} v_n,v_i)\\ &=a_{1j} T(v_1,v_i) + \dots + a_{nj} T(v_n,v_i)\tag{bilinearity}\\ &=a_{ij}\tag{$T(v_i,v_j)=\delta_{ij}$}\\ \end{align*}

By the same logic,

\begin{align*} T(A v_j,v_i)&=T(v_j,A v_i)\\ &=T(v_j,\sum_{k=1}^n a_{ki} v_k)\\ &=T(v_j,a_{1i} v_1)+\dots+T(v_j,a_{ni} v_n)\\ &=a_{1i} T(v_j,v_1)+\dots+a_{ni} T(v_j,v_n)\\ &= a_{ji}\\ \end{align*}

By hypothesis, $T(A v_j,v_i)=T(v_j,A v_i)$, therefore $a_{ij}=T(A v_j,v_i)=T(v_j,T v_i)=a_{ji}$.

I had this other idea though, that since $T$ is an inner product, its matrix is positive definite.

$T(x,f(y))=T(f(x),y)$ in matrix notation is $x^T T A y=(A x)^T T y$

\begin{align*} x^T T A y &= (A x)^T T y\\ &=x^T A^T T y\\ TA &= A^T T\\ (TA)^T &= (A^T T)^T\\ A^T T^T &= T^T A\\ TA &= A^T T^T\tag{T is symmetric}\\ &= (TA)^T\tag{transpose of matrix product}\\ \end{align*}

This is where I got stuck. We know that $T$ and $TA$ are both symmetric matrices. Clearly $T^{-1}$ is symmetric. If it can be shown that $T^{-1}$ and $AT$ commute, that would show it.

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With respect to your orthonormal basis, the matrix of $T$ is the identity matrix. So yes, it does commute with anything you want ;-) –  Alex B. Jan 21 '11 at 9:02
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2 Answers

up vote 11 down vote accepted

It's not true in general, e.g.

$A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ $B = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$

$AB = \begin{pmatrix} 4 & 3 \\ 3 & 2 \end{pmatrix}$

(with thanks to Rahul for formatting help)

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I did some numerical search for higher dimensions:

$n=3:$

$\left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right).\left( \begin{array}{ccc} -383 & 13 & -13 \\ -36 & -445 & -36 \\ -13 & 13 & -383 \end{array} \right)=\left( \begin{array}{ccc} -419 & -432 & -49 \\ -432 & -419 & -432 \\ -49 & -432 & -419 \end{array} \right)$

$n=4:$

$\left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{array} \right).\left( \begin{array}{cccc} -383 & 13 & -36 & -23 \\ 85 & -360 & 49 & -49 \\ -49 & 49 & -360 & 85 \\ -23 & -36 & 13 & -383 \end{array} \right)=\left( \begin{array}{cccc} -298 & -347 & 13 & -72 \\ -347 & -298 & -347 & 13 \\ 13 & -347 & -298 & -347 \\ -72 & 13 & -347 & -298 \end{array} \right)$

$n=5:$

$\left( \begin{array}{ccccc} 1 & 1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 0 & 1 & 3 & 1 & 0 \\ 0 & 0 & 1 & 4 & 1 \\ 0 & 0 & 0 & 1 & 5 \end{array} \right).\left( \begin{array}{ccccc} -7526 & 3158 & -9379 & 7340 & -8405 \\ 5216 & -3477 & 6079 & -6570 & 6359 \\ -3225 & 1486 & -3098 & 2500 & -3543 \\ 1159 & -1300 & 905 & -1249 & 970 \\ -641 & 414 & -841 & 186 & -656 \end{array} \right)=$

$=\left( \begin{array}{ccccc} -2310 & -319 & -3300 & 770 & -2046 \\ -319 & -2310 & -319 & -3300 & 770 \\ -3300 & -319 & -2310 & -319 & -3300 \\ 770 & -3300 & -319 & -2310 & -319 \\ -2046 & 770 & -3300 & -319 & -2310 \end{array} \right)$

All matrices have full rank. However, for high dimensions they are quite ugly.

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