# Every compact self-adjoint (linear) operator is injective (FALSE)

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?

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

The last step is wrong : since $$\|\phi\|,\|T\|$$ and $$\|\psi_\Lambda\|$$ are positive, the inequation : $$-\|\phi\|\|T\|\|\psi_\Lambda\| \leq 0$$ is a triviality.
• 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 $n \to \infty$" Jun 14, 2021 at 16:24
• 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 Jun 14, 2021 at 16:27
• The results states that $(\ker T)^\perp$ admits an orthonormal basis of eigenvectors. This gives no information on $\ker T$. Jun 14, 2021 at 16:28