# $U,T$ are self-adjoint with $TU=UT$, then there is orthonormal basis of eigenvectors of $T$ and $U$.

Let $V$ be a real finite-dimensional inner product space and $T,U$ be linear, self-adjoint operators on $V$ such that $TU=UT$. Prove that there is an orthonormal basis $\beta$ for $V$ consisting of eigenvectors of both $U$ and $T$.

I know that for any eigenvalues $\lambda$ of $T$, $E_\lambda$ and $E_\lambda^\perp$ are both $T$-,$U$-invariant. Also since $T$ is self-adjoint, there is an orthonormal basis consisting of egienvectors of $T$, but how to show that they are also eigenvectors of $U$?

• I think your question's title first words should be "$\;U,\,T\;$ are ..." . – DonAntonio Nov 13 '16 at 9:17
• – Math1000 Nov 13 '16 at 13:17

Since $T$ is self-adjoint (and thus diagonalizable by an orthonormal basis), we have $V=\bigoplus_i E_{\lambda_i}$ and moreover eigenvectors (of $T$) of different eigenvalues are orthogonal.
Since each eigenspace $E_{\lambda_i}$ is also $U$-invariant, this means we may write $U$ as $U = \oplus_i U_{E_{\lambda_i}}$. That is, the matrix of $U$ [in any basis for $V$ obtained by combining bases for each eigenspace $E_{\lambda_i}$ of $T$] is block diagonal, with one block per eigenspace of $T$.
We will show that each "block" $U_{E_{\lambda_i}}$ is diagonalizable. This is simple: $U_{E_{\lambda_i}}$ (as a map on $E_{\lambda_i}$) is still self-adjoint, so it is diagonalizable by some orthogonal basis $\beta_i$ on $E_{\lambda_i}$. Under this basis, $T_{E_{\lambda_i}}$ is still diagonalizable since $T_{E_{\lambda_I}} = \lambda_i I$ by definition of an eigenspace. Thus, we have found an orthogonal basis $\beta_i$ for each eigenspace $E_{\lambda_i}$ of $T$ such that both $T_{E_{\lambda_i}}$ and $U_{E_{\lambda_i}}$ are diagonalizable with respect to $\beta_i$. Using the fact that the $E_{\lambda_i}$ are "orthogonal" to each other, we see that combining the $\beta_i$ into one basis $\beta$ for $V$ gives an orthogonal basis such that $T$ and $U$ are both diagonalizable with respect to $\beta$.