Spectral decomposition of symmetric matrices

5.1. Spectral decomposition of symmetric matrices#

In this section the spectral decomposition of symmetric matrices (or Hermitian, if defined on the complex field), \(\mathbf{A} = \mathbf{A}^*\), is discussed.

Theorem 5.1 (Spectral theorem of a symmetric matrix)

A symmetric matrix has a spectral decomposition, with a complete set of unit orthogonal vectors and real eigenvalues,

\[\mathbf{A} = \mathbf{R} \mathbf{S} \mathbf{R}^* \ ,\]

with \(\mathbf{R} = \left[ \ \mathbf{r}_1 \ | \ \dots \ | \ \mathbf{r}_n \ \right]\), and \(\mathbf{r}_i^* \mathbf{r}_k = \delta_{ik}\), and \(\mathbf{S} = \text{diag}\{ s_i \}\), \(s_i \in \mathbb{R}\).

\[\begin{split}\begin{aligned} \mathbf{A} & = \mathbf{R} \mathbf{S} \mathbf{R}^* = \\ & = \begin{bmatrix} \mathbf{r}_1 & \dots & \mathbf{r}_n \end{bmatrix} \text{diag}\{ s_i \} \begin{bmatrix} \mathbf{r}_1^* \\ \dots \\ \mathbf{r}_n^* \end{bmatrix} = \\ & = \begin{bmatrix} s_1 \mathbf{r}_1 & \dots & s_n \mathbf{r}_n \end{bmatrix} \begin{bmatrix} \mathbf{r}_1^* \\ \dots \\ \mathbf{r}_n^* \end{bmatrix} = \\ & = s_1 \mathbf{r}_1 \mathbf{r}_1^* + \dots + s_n \mathbf{r}_n \mathbf{r}_n^* \ . \end{aligned}\end{split}\]

Properties.

Left and right eigenvectors are the same. This immediately follows from the symmetry of the matrix and the definition of the left and right eigenvectors

\[\begin{split}\begin{aligned} \mathbf{A} \mathbf{r}_k & = s_k \mathbf{r}_k \\ \mathbf{l}_j^* \mathbf{A} & = s^*_j \mathbf{l}_i^* \quad \rightarrow \quad \mathbf{A} \mathbf{l}_j = s_j \mathbf{l}_j \ , \end{aligned}\end{split}\]
Eigenvalues are real, as

\[0 = \mathbf{l}^*_k \left( \mathbf{A} \mathbf{r}_k \right) - \left( \mathbf{l}_k^* \mathbf{A} \right) \mathbf{r}_k = \left( s_k - s^*_k \right) \mathbf{l}^*_k \mathbf{r}_k = 2 \, \text{im}\{ s_k \} \underbrace{\mathbf{l}^*_k \mathbf{r}_k}_{ = | \mathbf{r}_k |^2} \ ,\]

and thus \(\text{im} \{ s_k \} = 0\).
Eigenvectors with different eigenvalues are orthogonal, as

\[0 = \mathbf{l}^*_j \left( \mathbf{A} \mathbf{r}_k \right) - \left( \mathbf{l}_j^* \mathbf{A} \right) \mathbf{r}_k = \left( s_k - s^*_j \right) \mathbf{l}^*_j \mathbf{r}_k \ ,\]

and thus \(\mathbf{l}^*_j \mathbf{r}_k = 0\) for \(j \ne k\), if \(s_j \ne s_k\). Left and right eigenvectors are orthogonal w.r.t. matrix \(\mathbf{A}\) as well.