Operator Diagonalizable iff Basis of Eigenvectors

Theorem

Let $\HH$ be a Hilbert space.

Let $A: \HH \to \HH$ be a linear operator on $\HH$.

Then $A$ is diagonalizable if and only if there exists a basis $E$ of $\HH$, consisting of eigenvectors for $A$.

Proof

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Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.7.4$