Let $V$ be a finite dimensional $\mathbb R$-vector space and let $T:V\rightarrow V$ be an self-adjoint operator such that $\text{trace}(T)=0$. Show that there exists an orthonormal basis $B$ such that every element of the diagonal of $[T]_B$ is $0$.

  • $\begingroup$ Perhaps I am missing something, but do you already know that a symmetric operator is orthogonally diagonalizable? Because if you do then we're done... $\endgroup$ – DonAntonio Nov 20 '16 at 22:12

I'm not aware of a completely elementary proof of this result (not saying that there isn't).

Consider the numerical range $$W(A)=\{\langle Ax,x\rangle:\ \|x\|=1\}.$$ They key fact is the Toeplitz-Hausdorff Theorem, that says that $W(A)$ is convex.

Since $\text{Tr}(A)=0$, we can write (for an orthonormal basis $\{e_j\}$) $$ 0=\sum_{j=1}^n\frac1n\,\langle Ae_j,e_j\rangle\in W(A). $$ It follows that there exists $x\in V$, with $\|x\|=1$, such that $\langle Ax,x\rangle=0$. Now extend $\{x\}$ to an orthonormal basis $\{x_j\}$, where $x_1=x$. Then $$ 0=\text{Tr}(A)=\sum_{j=1}^n\langle Ax_j,x_j\rangle=\sum_{j=2}^n\langle Ax_j,x_j\rangle. $$ The above equality shows that if $V_2=\text{span}\{x_2,\ldots,x_n\}\subset V$ then $A_2=P_{V_2}AP_{V_2}$, as an operator on $V_2$, has trace zero. That is, $$ A=\begin{bmatrix}0&*\\ *&A_2\end{bmatrix} $$ with $\text{Tr}(A_2)=0$. So now the result follows by induction.


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