Exercise 1751Linear algebra › Inner product spacesL1Added 3 Nov 2026

Self-adjoint operators and symmetric matrices

#self-adjoint#symmetric matrix#eigenvalues

Statement

Show that a linear map $f:\mathbb{R}^n\to\mathbb{R}^n$ is self-adjoint (i.e. $f=f^*$ ) iff its matrix in an orthonormal basis is symmetric.

Hint

Use $f^*$ has matrix $M^T$ .

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Solution

If $M$ is symmetric. Then the matrix of $f^*$ is $M^T=M$ , so $f^*=f$ .

If $f$ is self-adjoint. Then the matrix of $f^*$ equals the matrix of $f$ . But the matrix of $f^*$ is $M^T$ , so $M=M^T$ .

Conclusion. In an orthonormal basis, $f$ is self-adjoint iff $M$ is symmetric.

Consequence. Self-adjoint operators have real eigenvalues, and eigenvectors for distinct eigenvalues are orthogonal (spectral theorem for symmetric matrices).

Self-adjoint operators and symmetric matrices

Statement

Solution

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