My purpose is to provide a proof, without using the spectral theorem.

Let $T: \scr H \to H$ be a linear compact self-adjoint operator.

$T$ self-adjoint implies:

  1. $ \langle \phi, T \psi \rangle = \langle T\phi, \psi \rangle = \overline{\langle \phi, T \psi \rangle}$, which implies $\langle \phi, T \psi \rangle \in \mathbb R$.
  2. $- \| \phi\| \|T\| \|\psi\| \le \langle \phi, T \psi \rangle \le \| \phi\| \|T\| \|\psi\|$

If in addition $T$ is compact, then its point spectrum $\sigma_p(T)\equiv \{ \lambda \in \mathbb C \ |\ T\psi = \lambda \psi$, $ \ $ for some $\lambda$-eigenvector $\psi \in \scr H$$\}$ is real, non-void.

In particular $\exists\ \Lambda \in \sigma_p(T)$ that verifies $|\Lambda| = \|T\| = \underset {\sigma_p(T)}{\max}|\lambda|$. Let's call the corresponding non-null $\Lambda$-eigenvector $\psi_\Lambda$.

remark. 1. and 2. hold $\forall\ \phi, \psi \in \scr H$. If we take $\phi \in \ker(T)$ and $\psi = \psi_\Lambda$:

$- \| \phi\| \|T\| \|\psi_\Lambda\| \le \langle \phi, T \psi_\Lambda \rangle = \langle T\phi, \psi_\Lambda \rangle = 0 $

Since $\|T\|$ and $\|\psi_\Lambda\|$ are not zero, the only possibility left is $\|\phi\| =0 \implies\phi ={\bf 0} \implies \ker(T) \equiv \{\bf 0\} $.

$T$ is hence injective. CVD(?)

Is this proof correct?

  • 3
    $\begingroup$ This is not true. The $0$ operator or any self-adjoint finite rank operator satisfies the prompt but not the conclusion. $\endgroup$ Jun 14, 2021 at 16:01

1 Answer 1


The last step is wrong : since $\|\phi\|,\|T\|$ and $\|\psi_\Lambda\|$ are positive, the inequation : $$-\|\phi\|\|T\|\|\psi_\Lambda\| \leq 0$$ is a triviality.

The result itself is wrong : a finite rank orthogonal projector is compact, bounded and self-adjoint, but is not injective.

  • $\begingroup$ Yes, I forgot the minus sign. Thanks. $\endgroup$
    – ric.san
    Jun 14, 2021 at 16:10
  • $\begingroup$ But then... what about the spectral theorem? "For every compact self-adjoint operator $T$ on a Hilbert space $H$, there exists an orthonormal basis of $H$ consisting of eigenvectors of $T$. More specifically, the orthogonal complement of $\ker T$ admits either a finite orthonormal basis of eigenvectors of $T$, or a countably infinite orthonormal basis $ \{ \psi_n \}_{n \in \mathbb N}$ of eigenvectors of $T$, with corresponding eigenvalues $ \{ \lambda_n \}_{n \in \mathbb N} \in \mathbb R$, such that $\lambda_n \to 0$ as $n \to \infty$" $\endgroup$
    – ric.san
    Jun 14, 2021 at 16:24
  • $\begingroup$ For an orthonormal basis to be such, ${\rm span}\{ \psi_n \} = \{ \ker T\}^\perp$ has to be dense in $H$. That means $\ker T$ has to be trivial $\endgroup$
    – ric.san
    Jun 14, 2021 at 16:27
  • $\begingroup$ The results states that $(\ker T)^\perp$ admits an orthonormal basis of eigenvectors. This gives no information on $\ker T$. $\endgroup$ Jun 14, 2021 at 16:28
  • 1
    $\begingroup$ No, it's an orthonormal basis of the subspace $\endgroup$ Jun 14, 2021 at 16:32

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