Spectral decomposition

Task number: 4472

Show that every Hermitian matrix \(\boldsymbol A\) of order \(n\) can be written as \(\boldsymbol A=\lambda_1\boldsymbol v_1\boldsymbol v_1^{\mathsf H}+\lambda_2\boldsymbol v_2\boldsymbol v_2^{\mathsf H}+\cdots+\lambda_n\boldsymbol v_n\boldsymbol v_n^{\mathsf H}\), where \(\boldsymbol v_1,\dots,\boldsymbol v_n\) are suitably chosen eigenvectors corresponding to (not necessarily different) eigenvalues \(\lambda_1,\dots,\lambda_n\).

Solution
Any Hermitian matrix can be written as \(\boldsymbol A=\boldsymbol {RDR}^{\mathsf H}\), where \(\boldsymbol D\) is diagonal with eigenvalues on the diagonal and \(\boldsymbol R\) is unitary, i.e. \(\boldsymbol R^{\mathsf H}=\boldsymbol R^{-1}\).

The columns of \(\boldsymbol R\) are the eigenvectors of the matrix \(\boldsymbol A\), and we choose these for \(\boldsymbol v_1,\dots,\boldsymbol v_n\). The columns of the product \(\boldsymbol {RD}\) are then \(\lambda_1\boldsymbol v_1,\dots,\lambda_n\boldsymbol v_n\).

Vectors \(\boldsymbol v_1^{\mathsf H},\dots,\boldsymbol v_n^{\mathsf H}\) are rows of \(\boldsymbol R^{\mathsf H}\). Now it suffices to interpret the outer product in the expression \((\boldsymbol {RD})\boldsymbol R^{\mathsf H}\) as the sum of \(n\) matrices \(\sum\limits_{i=1}^n (\lambda_i\boldsymbol v_i)\boldsymbol v_i^{\mathsf H}\), which holds indeed in general for any product of two matrices (see task Column products).

Spectral decomposition

Task number: 4472

Solution