5.3: Symmetric and Hermitian Matrices
When a real matrix \(A\) is equal to its transpose, \(A^{T}=A\) , we say that the matrix is symmetric. When a complex matrix \(A\) is equal to its conjugate transpose, \(\mathrm{A}^{+}=\mathrm{A}\) , we say that the matrix is Hermitian.
One of the reasons symmetric and Hermitian matrices are important is because their eigenvalues are real and their eigenvectors are orthogonal. Let \(\lambda_{i}\) and \(\lambda_{j}\) be eigenvalues and \(x_{i}\) and \(x_{j}\) eigenvectors of the possibly complex matrix A. We have
\[\mathrm{A} x_{i}=\lambda_{i} x_{i}, \quad \mathrm{~A} x_{j}=\lambda_{j} x_{j} . \nonumber \]
Multiplying the first equation on the left by \(x_{j}^{\dagger}\) , and taking the conjugate transpose of the second equation and multiplying on the right by \(x_{i}\) , we obtain
\[x_{j}^{\dagger} \mathrm{A} x_{i}=\lambda_{i} x_{j}^{\dagger} x_{i}, \quad x_{j}^{\dagger} \mathrm{A}^{\dagger} x_{i}=\bar{\lambda}_{j} x_{j}^{\dagger} x_{i} . \nonumber \]
If \(A\) is Hermitian, then \(A^{+}=A\) , and subtracting the second equation from the first yields
\[\left(\lambda_{i}-\bar{\lambda}_{j}\right) x_{j}^{\dagger} x_{i}=0 \nonumber \]
If \(i=j\) , then since \(x_{i}^{\dagger} x_{i}>0\) , we have \(\bar{\lambda}_{i}=\lambda_{i}\) : all eigenvalues are real. If \(i \neq j\) and \(\lambda_{i} \neq \lambda_{j}\) , then \(x_{j}^{\dagger} x_{i}=0\) : eigenvectors with distinct eigenvalues are orthogonal. Usually, the eigenvectors are made orthonormal, and diagonalization makes use of real orthogonal or complex unitary matrices.
Example: Diagonalize the symmetric matrix
\[\mathrm{A}=\left(\begin{array}{ll} a & b \\ b & a \end{array}\right) \nonumber \]
The characteristic equation of \(\mathrm{A}\) is given by
\[(a-\lambda)^{2}=b^{2}, \nonumber \]
with real eigenvalues \(\lambda_{1}=a+b\) and \(\lambda_{2}=a-b\) . The eigenvector with eigenvalue \(\lambda_{1}\) satisfies \(-x_{1}+x_{2}=0\) , and the eigenvector with eigenvalue \(\lambda_{2}\) satisfies \(x_{1}+x_{2}=0\) . Normalizing the eigenvectors, we have
\[\lambda_{1}=a+b, \mathrm{X}_{1}=\frac{1}{\sqrt{2}}\left(\begin{array}{l} 1 \\ 1 \end{array}\right) ; \quad \lambda_{2}=a-b, \mathrm{X}_{2}=\frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ -1 \end{array}\right) \nonumber \]
Evidently, the eigenvectors are orthonormal. The diagonalization using \(\mathrm{A}=\mathrm{Q} \Lambda \mathrm{Q}^{-1}\) is given by
\[\left(\begin{array}{ll} a & b \\ b & a \end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right)\left(\begin{array}{cc} a+b & 0 \\ 0 & a-b \end{array}\right) \frac{1}{\sqrt{2}}\left(\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right), \nonumber \]
which can be verified directly by matrix multiplication. The matrix \(Q\) is a symmetric orthogonal matrix so that \(Q^{-1}=Q\) .