# T normal operator has splitting characteristic polynomial - eigenvectors form basis?

• Nov 24th 2009, 05:57 PM
Last_Singularity
T normal operator has splitting characteristic polynomial - eigenvectors form basis?
Question: $\displaystyle T$ is a normal operator on a finite-dimensional real inner product space $\displaystyle V$ with a characteristic polynomial that splits. Prove that $\displaystyle V$ has an orthonormal basis of eigenvectors of $\displaystyle T$.

By Schur's theorem, because the characteristic polynomial of $\displaystyle T$ splits, there exists an orthonormal basis $\displaystyle \beta$ for $\displaystyle V$ such that $\displaystyle [T]_{\beta}$ is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of $\displaystyle T$? Thanks!
• Nov 24th 2009, 06:57 PM
tonio
Quote:

Originally Posted by Last_Singularity
Question: $\displaystyle T$ is a normal operator on a finite-dimensional real inner product space $\displaystyle V$ with a characteristic polynomial that splits. Prove that $\displaystyle V$ has an orthonormal basis of eigenvectors of $\displaystyle T$.

By Schur's theorem, because the characteristic polynomial of $\displaystyle T$ splits, there exists an orthonormal basis $\displaystyle \beta$ for $\displaystyle V$ such that $\displaystyle [T]_{\beta}$ is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of $\displaystyle T$? Thanks!

Schur's theorem tells us that $\displaystyle U^{*}TU=A$ , where $\displaystyle A$ is upper triangular and $\displaystyle U$ is unitary.
As all the eigenvalues of T are real (otherwise its char. pol. wouldn't split over the reals!), T is self-adjoint or Hermitian (i.e., $\displaystyle T=T^{*}$), and then:

$\displaystyle A^{*}=\left(U^{*}TU\right)^{*}=U^{*}T^{*}U=U^{*}TU =A\Longrightarrow\,A=A^{*}$ , and since $\displaystyle A$ is upper triangular then $\displaystyle A^{*}$ is lower triangular, thus $\displaystyle A$ is in fact diagonal and we're done.

Tonio