Symmetric and Hermitian Matrices

Definitions and Entrywise Meaning

A real square matrix $A\in\mathbb{R}^{n\times n}$ is $symmetric$ when

A^T=A

Entry by entry, the same condition is $a_{ji}=a_{ij}$ . For example,

A=\begin{pmatrix}1&2&3\\2&4&5\\3&5&6\end{pmatrix}

The matrix $A$ is symmetric because its entries match across the main diagonal.

A complex square matrix $H\in\mathbb{C}^{n\times n}$ is $Hermitian$ when

H^H=H

Entry by entry, the Hermitian condition means $h_{ji}=\overline{h_{ij}}$ . An example with complex entries is

H=\begin{pmatrix}2&1+i\\1-i&3\end{pmatrix}

Every real symmetric matrix is Hermitian when viewed over $\mathbb{C}$ . The converse is false because a Hermitian matrix may have complex off-diagonal entries.

On the diagonal, the Hermitian condition gives $h_{ii}=\overline{h_{ii}}$ , so every diagonal entry is real.

Real and Complex Inner Products

The transpose and conjugate transpose must not be mixed. For $x\in\mathbb{R}^n$ ,

x^Tx=\sum_{k=1}^n x_k^2\geq0

For $x\in\mathbb{C}^n$ , $use the conjugate transpose$ :

x^Hx=\sum_{k=1}^n\overline{x_k}x_k=\sum_{k=1}^n|x_k|^2\geq0

In either field, the relevant expression equals zero exactly when $x=0$ . For a complex vector, $x^Tx$ need not even be real, which is why complex geometry uses $x^Hx$ .

Quadratic and Hermitian Forms

A real symmetric matrix defines the quadratic form

q:\mathbb{R}^n\to\mathbb{R}

A Hermitian matrix defines the real-valued Hermitian form

q:\mathbb{C}^n\to\mathbb{R}

To prove that $x^HHx$ is real, take its complex conjugate.

\overline{x^HHx}=(x^HHx)^H

A scalar equals its complex conjugate exactly when it is real. In the real case, the skew-symmetric part of any matrix contributes nothing to a quadratic form:

x^TBx=x^T\left(\frac{B+B^T}{2}\right)x

The skew-symmetric part contributes nothing because $x^T(B-B^T)x=0$ . Thus, $quadratic forms depend only on the symmetric part$ .

Why Hermitian Eigenvalues Are Real

Suppose $Hv=\lambda v$ with $v\neq0$ . $Hermitian symmetry$ gives

\lambda v^Hv=v^HHv

Since $v^Hv>0$ , it follows that $\lambda=\overline{\lambda}$ and therefore $\lambda\in\mathbb{R}$ . Real symmetric matrices inherit the same conclusion.

Why Distinct Eigenspaces Are Orthogonal

Let $Hv=\lambda v$ and $Hw=\mu w$ with $\lambda\neq\mu$ . Because $both eigenvalues are real$ ,

\lambda w^Hv=w^HHv

Hence $(\lambda-\mu)w^Hv=0$ , so $w^Hv=0$ . With a repeated eigenvalue, arbitrary eigenvectors inside the same eigenspace may fail to be orthogonal. Gram-Schmidt can produce an orthonormal basis within that eigenspace.

Spectral Decomposition and Quadratic Geometry

The spectral theorem collects these facts into a complete decomposition:

A=A^T\in\mathbb{R}^{n\times n}:\qquad A=Q\Lambda Q^T

The columns of $Q$ or $U$ form an $orthonormal eigenbasis$ , and every diagonal entry of $\Lambda$ is real. If $u_1,\ldots,u_n$ are the columns of $U$ , then

x^HHx=\sum_{j=1}^n\lambda_j|u_j^Hx|^2

This formula shows immediately that $H$ is positive definite exactly when every eigenvalue is positive.

Worked Hermitian Diagonalization

For the earlier matrix

H=\begin{pmatrix}2&1+i\\1-i&3\end{pmatrix}

For the matrix $H$ , the characteristic polynomial is

\begin{aligned} \det(H-\lambda I) &=(2-\lambda)(3-\lambda)-2 \\ &=(\lambda-1)(\lambda-4) \end{aligned}

$Normalized eigenvectors$ for $\lambda_1=1$ and $\lambda_2=4$ are

u_1=\frac1{\sqrt3}\begin{pmatrix}-(1+i)\\1\end{pmatrix}

They satisfy $u_1^Hu_2=0$ . With $U=(u_1\quad u_2)$ ,

U^HHU=\operatorname{diag}(1,4)

Check the orthogonality and the first eigenvector directly. The normalizing factor stays outside each product.

u_1^Hu_2=\frac{-(1-i)(1+i)+2}{\sqrt{18}}=\frac{-2+2}{\sqrt{18}}=0