symmetric matrix

Symmetric Matrix

A matrix $A$ is a symmetric if it does not change under transpose: $A=A^T$, or equivalently $Q_{ij}=Q_{ji}\forall 1,\dots ,n$.

For complex matrices with $A=A^{\ast }$ we call $A$ hermitian / self-adjoint,.

$A+A^T$ is always symmetric, even if $A$ is not.

$$\begin{array}{rl}(A+A^T{)}^T& =A^T+(A^T{)}^T=A^T+A\\ & =A+A^T=(A+A^T{)}^T⟹\text{symmetric}\end{array}$$

Any matrix can be decomposed into a symmetric and an antisymmetric part:

$$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$$

The first part is symmetric, the second part is antisymmetric, i.e. $(A-A^T{)}^T=-(A-A^T)$.

Spectral theorem

Every symmetric matrix has an eigendecomposition with an orthonormal basis of eigenvectors.

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Let $A\in {\mathbb{R}}^{n\times n}$ be a symmetric matrix. Then, all eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$ of $A$ are real, and there exists an orthonormal basis of ${\mathbb{R}}^n$ consisting of eigenvectors of $A$ (aka eigenbasis).
Then:

$$A=A^T⟹A=Q\Lambda Q^T$$

$Q$ … orthogonal matrix ¹² ($Q^{-1}=Q^T$) whose columns are the eigenvectors of $A$
$\Lambda =\text{diag}({\lambda }_1,\dots ,{\lambda }_n)$ … diagonal matrix with the eigenvalues of $A$ on the diagonal.
→ For symmetric matrices, this is a pure scaling in the orthogonal directions of the eigenvectors or principal axes - the natural directions along which $A$ acts purely by scaling. Each eigenvector direction $v_k$ is scaled by its eigenvalue ${\lambda }_k$ without any rotation or shearing.
→ Eigendecomposition and SVD coincide for symmetric matrices.

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