Matrix of self-adjoint operator such that every element of the diagonal is $0$.

Question

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 $0$.

Answer 1

(thanks to Miguel Wazowski, here is a fairly elementary proof)

Since $A$ is selfadjoint, it admits an orthonormal basis of eigenvectors $v_1,\ldots,v_n$. Let $x=c\,\sum_jv_j$, with $c$ chosen so that $\|x\|=1$. We have $$ \langle Ax,x\rangle=c^2\,\sum_{k,j}\langle Av_j,v_k\rangle=c^2\sum_{k,j}\lambda_j\langle v_j,v_k\rangle=c^2\,\sum_j\lambda_j=c^2\operatorname{Tr}(A)=0. $$ 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.