2
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.