# Diagonal element of traceless hermitian matrix?

In physics, we are familiar with a set of traceless hermitian matrices named Pauli matrices: {\displaystyle {\begin{aligned}\sigma _{1}=\sigma _{\mathrm {x} }&={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\\\sigma _{2}=\sigma _{\mathrm {y} }&={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\\\sigma _{3}=\sigma _{\mathrm {z} }&={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\\\end{aligned}}} I notice that the above matrices are written in basis $$\left( \begin{array}{c} 1\\ 0\\ \end{array} \right)$$ and $$\left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$$, but if we change basis into $$\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1\\ 1\\ \end{array} \right)$$ and $$\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1\\ -1\\ \end{array} \right)$$, then $$\sigma_z$$ become $$\frac{1}{2}\left( \begin{array}{c} 1\\ 1\\ \end{array} \right) \left( \begin{matrix} 1& 1\\ \end{matrix} \right) -\frac{1}{2}\left( \begin{array}{c} 1\\ -1\\ \end{array} \right) \left( \begin{matrix} 1& -1\\ \end{matrix} \right) =\left( \begin{matrix} 0& 1\\ 1& 0\\ \end{matrix} \right) ,$$showing that the diagonal elements vanish.

My question is, for a single traceless hermitian matrix $$H$$, can we always find a unitary $$u$$ such that diagonal element of $$u^{\dagger}Hu$$ are all zeros? Furthermore, can the diagonal elements run over all possibilities as long as they satisfy sum up to zero?

• Do you mean to find a matrix $u$ such that $u^{-1}\sigma_iu$ has a zero diagonal for every $i\in\{1,2,3\}$? Do you require $u$ to be unitary? Commented Apr 26, 2022 at 13:20
• @user1551 No, I only need this work for a single traceless hermitian matrix. I'm sorry for the ambiguity. I just use pauli matrices to show my original meaning... Commented Apr 26, 2022 at 13:41

Yes. Let $$H$$ be any $$n\times n$$ traceless Hermitian matrix and $$UDU^\ast$$ be its unitary diagonalisation. Let $$Q$$ be a real orthogonal matrix whose last column is $$\frac{1}{\sqrt{n}}(1,1,\ldots,1)^T$$. The last diagonal element of $$H':=Q^TU^\ast HUQ=Q^TDQ$$ is then the mean of all diagonal elements of $$H$$, which is zero. Since $$H'$$ is real symmetric, we can do the similar to the leading principal $$(n-1)$$-rowed submatrix of $$H'$$ and continue recursively to obtain ultimately a real symmetric matrix with a zero diagonal.