I am trying to prove that the eigenvectors of a self-adjoint operator $T$ defined on a Hilbert space $\mathcal{H}$ are orthogonal.

I did the following:

For any eigenvectors $\mathbf{x}$ and $\mathbf{y}$ of $T$ corresponding to distinct eigenvalues $\lambda$ and $\mu$.

Then $$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle T\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},T^\star\mathbf{y}\rangle = \langle\mathbf{x},T\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$ Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$.

My question is how to relax the assumption that the corresponding eigenvalues are distinct; i.e. what should I do if the eigenvalues are not distinct?

  • 3
    $\begingroup$ You can't do anything if not distinct. Consider the identity in finite-dimensional spaces. You could Gram-Schmidt the eigenspace (or use Zorn's Lemma to get an orthonormal set). $\endgroup$ – max_zorn Apr 8 '18 at 23:12

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