Spectral Theorem

Normal Matrices

The complex spectral theorem states that a matrix can be diagonalized using an orthonormal basis of eigenvectors $exactly when it is normal$ . In matrix form,

$A$ is normal if and only if it has this decomposition with unitary $U$ and diagonal $\Lambda$ :

A=U\Lambda U^H

In this form, $U$ is unitary and $\Lambda$ is diagonal. The columns of $U$ form an orthonormal basis of eigenvectors. Because $U$ is unitary, this change of basis preserves lengths and angles. In the real symmetric case, $U$ can be chosen real and orthogonal, so $U^{-1}=U^T$ .

A complex matrix $A \in \mathbb{C}^{n \times n}$ is called normal if it satisfies the commutativity condition with its conjugate transpose:

A \cdot A^H = A^H \cdot A

This algebraic condition makes eigenspaces for distinct eigenvalues orthogonal. Over $\mathbb{C}$ , it also allows a unitary diagonalization.

Special Properties of Eigenspaces of Normal Matrices

For a normal matrix, the null space of the matrix equals the null space of its conjugate transpose.

The null space (kernel) is the set of all vectors $x$ satisfying $Ax=0$ . Normality gives the following $norm identity$ for every $x$ :

\lVert Ax\rVert_2^2=x^HA^HAx

$Ax=0$ if and only if $A^Hx=0$ , so

\ker A=\ker A^H

Apply the same null-space identity to the normal matrix $A-\lambda I$ . For every eigenvalue $\lambda \in \mathbb{C}$ , this gives the corresponding eigenspace relation:

E_A(\lambda)=E_{A^H}(\overline{\lambda})

If $Av=\lambda v$ , then $A^H v=\overline{\lambda}v$ . The eigenvector is shared, while the eigenvalue is complex-conjugated.

This identity also proves orthogonality. If $Av=\lambda v$ and $Aw=\mu w$ with $\lambda\neq\mu$ , then

\overline{\lambda}v^Hw=(Av)^Hw=v^HA^Hw

Hermitian and Unitary Matrices as Examples of Normal Matrices

A Hermitian matrix equals its own adjoint, and a unitary matrix has an inverse equal to its adjoint, so in both cases the two sides of the normality condition collapse into the $same product$ .

Hermitian Matrices and Real Eigenvalues

For a Hermitian matrix, $A^H=A$ , so both sides of the normality condition equal $A^2$ .

Eigenvalues of Hermitian matrices are $always real$ . If $Av=\lambda v$ with $v\neq0$ , then Hermitian symmetry gives:

\begin{aligned} \lambda v^H v &= v^HAv \\ &= v^HA^Hv \\ &= (Av)^Hv \\ &= \overline{\lambda}v^Hv \\ &\Rightarrow \lambda \\ &=\overline{\lambda} \end{aligned}

The equality $\lambda=\overline{\lambda}$ says exactly that $\lambda$ is real.

Worked Hermitian Example

Consider

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

The conjugate transpose returns the same matrix, so $H^H=H$ . Its characteristic polynomial is $(2-\lambda)^2-1$ , which gives the real eigenvalues $1$ and $3$ . Corresponding normalized eigenvectors are

\begin{aligned} &u_1=\frac1{\sqrt2}\begin{pmatrix}-i\\1\end{pmatrix}, \\ &u_3=\frac1{\sqrt2}\begin{pmatrix}i\\1\end{pmatrix} \end{aligned}

The vectors are $orthonormal because$ $u_1^Hu_3=0$ and both have norm $1$ . This is the spectral theorem in a concrete $2\times2$ case.

u_1^Hu_3=\frac{(i)(i)+1}{2}=0,\qquad Hu_1=\frac1{\sqrt2}\begin{pmatrix}-2i+i\\(-i)(-i)+2\end{pmatrix}=u_1

Unitary Matrices and Eigenvalues on the Unit Circle

Unitary matrices have the property $A^H = A^{-1}$ . To prove that unitary matrices are also normal, we substitute into the definition and get $A \cdot A^H = A \cdot A^{-1} = I = A^{-1} \cdot A = A^H \cdot A$ .

Eigenvalues of unitary matrices have magnitude $1$ , so they lie on the $unit circle$ in the complex plane. If $Av=\lambda v$ with $v\neq0$ , unitarity preserves the norm:

\lVert v\rVert^2 = \lVert Av\rVert^2 = \lVert \lambda v\rVert^2 = |\lambda|^2\lVert v\rVert^2

Every eigenvalue of a unitary matrix has modulus $1$ . Geometrically, unitary transformations preserve vector lengths, and their eigenvalues lie on the unit circle.