# The eigenvectors of a normal matrix are an orthonormal basis.

Put content here**Theorem.** Every normal matrix has an orthonormal basis of eigenvectors.
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If $A$ is normal, then there exist orthonormal vectors $v_1, \ldots, v_n$ and eigenvalues $\lambda_1, \ldots, \lambda_n$ such that $Av_i = \lambda_i v_i$.
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**Proof.** By the Principal Axis Theorem, $A = UDU^*$ where $U$ is unitary and $D$ is diagonal. The columns of $U$ are orthonormal eigenvectors of $A$.
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**Key property.** For a normal matrix, eigenvectors corresponding to different eigenvalues are orthogonal:
If $Av = \lambda v$ and $Aw = \mu w$ with $\lambda \neq \mu$, then $\langle v, w \rangle = 0$.
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**Example.** The unitary matrix $U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is normal with orthonormal eigenvectors $\frac{1}{\sqrt{2}}(1,1)^T$ and $\frac{1}{\sqrt{2}}(1,-1)^T$.

# Parents

* Normal matrices