Spectral Theorem for Real Normal Matrices

When a Real Normal Matrix Has a Real Eigenbasis

For $A\in\mathbb{R}^{n\times n}$ , the following statements are equivalent:

There is an orthonormal basis of $\mathbb{R}^n$ consisting of eigenvectors of $A$ .
The matrix is normal and its characteristic polynomial $\chi_A(t)$ splits completely into linear factors over $\mathbb{R}$ .
There are a real orthogonal matrix $Q$ and a real diagonal matrix $\Lambda$ such that $A=Q\Lambda Q^T$ .
The matrix is symmetric: $A^T=A$ .

The equivalence explains the extra condition compared with complex normal matrices. A complex normal matrix always has a unitary eigenbasis, but $that basis and its eigenvalues need not be real$ .

The proof links the four statements directly. An orthonormal eigenbasis gives the columns of $Q$ . From $A=Q\Lambda Q^T$ with real diagonal $\Lambda$ , we obtain $A^T=A$ . Conversely, the real spectral theorem orthogonally diagonalizes every real symmetric matrix. Finally, if a real normal matrix has only real eigenvalues, its complex unitary diagonalization satisfies $A^H=A$ . Because $A$ is real, this is exactly $A^T=A$ .

Matrix class	Field	Guaranteed spectral form
Complex normal	$\mathbb{C}$	Unitary diagonalization. Eigenvalues may be complex
Complex Hermitian	$\mathbb{C}$	Unitary diagonalization with real eigenvalues
Real symmetric	$\mathbb{R}$	Orthogonal diagonalization with real eigenvalues
Real orthogonal	$\mathbb{R}$	In general, the spectral form contains real rotation blocks. A real diagonal form exists only when all eigenvalues are real

A Normal Matrix without a Real Eigenbasis

Consider the rotation through an angle $\alpha$ that is not an integer multiple of $\pi$ :

R_\alpha=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}

The matrix is orthogonal, so $R_\alpha^TR_\alpha=I=R_\alpha R_\alpha^T$ and it is normal. Its characteristic polynomial is

\chi_{R_\alpha}(t)=(t-\cos\alpha)^2+\sin^2\alpha

Its discriminant is

\Delta=4\cos^2\alpha-4=-4\sin^2\alpha<0

Because $\alpha$ is not an integer multiple of $\pi$ , $\sin\alpha\neq0$ . Thus the characteristic polynomial has $no real root$ , and $R_\alpha$ has no real eigenvector basis. Over $\mathbb{C}$ , the eigenvalues are $e^{i\alpha}$ and $e^{-i\alpha}$ .

At $\alpha=\pi/2$ , the blue vector $v=(1,1)^T$ becomes the orange vector $Rv=(-1,1)^T$ . Their dot product is zero. More generally, $(x,y)\cdot(-y,x)=0$ , so every nonzero real vector turns by 90°, rather than staying on its original line.

A quarter-turn changes every real direction

The blue vector and its orange image are perpendicular. A quarter-turn preserves no direction line of a nonzero real vector.

$x$: $[-1, 1]$
$y$: $[0, 1]$

A Diagonalizable Matrix That Is Not Normal

Consider the following matrix over $\mathbb{R}$ :

A=\begin{pmatrix}1&3\\0&2\end{pmatrix}

Its eigenvalues $1$ and $2$ are distinct, so the matrix is diagonalizable. The corresponding eigenspaces are

E_A(1)=\ker(A-I)=\operatorname{span}\begin{pmatrix}1\\0\end{pmatrix}

Using these eigenvectors as columns gives an explicit diagonalization:

S=\begin{pmatrix}1&3\\0&1\end{pmatrix},\qquad D=\begin{pmatrix}1&0\\0&2\end{pmatrix}

However, the two eigenvectors are not orthogonal because

\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=3\neq0

Direct multiplication also shows that $A$ is not normal:

AA^T=\begin{pmatrix}10&6\\6&4\end{pmatrix}

This example separates the two ideas: ordinary diagonalization permits any invertible eigenvector matrix $S$ , while $orthogonal diagonalization requires an orthonormal eigenbasis$ and therefore a real symmetric matrix.

When a Real Normal Matrix Has a Real Eigenbasis

For $A\in\mathbb{R}^{n\times n}$ , the following statements are equivalent:

There is an orthonormal basis of $\mathbb{R}^n$ consisting of eigenvectors of $A$ .
The matrix is normal and its characteristic polynomial $\chi_A(t)$ splits completely into linear factors over $\mathbb{R}$ .
There are a real orthogonal matrix $Q$ and a real diagonal matrix $\Lambda$ such that $A=Q\Lambda Q^T$ .
The matrix is symmetric: $A^T=A$ .

The equivalence explains the extra condition compared with complex normal matrices. A complex normal matrix always has a unitary eigenbasis, but $that basis and its eigenvalues need not be real$ .

Matrix class	Field	Guaranteed spectral form
Complex normal	$\mathbb{C}$	Unitary diagonalization. Eigenvalues may be complex
Complex Hermitian	$\mathbb{C}$	Unitary diagonalization with real eigenvalues
Real symmetric	$\mathbb{R}$	Orthogonal diagonalization with real eigenvalues
Real orthogonal	$\mathbb{R}$	In general, the spectral form contains real rotation blocks. A real diagonal form exists only when all eigenvalues are real

A Normal Matrix without a Real Eigenbasis

Consider the rotation through an angle $\alpha$ that is not an integer multiple of $\pi$ :

R_\alpha=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}

The matrix is orthogonal, so $R_\alpha^TR_\alpha=I=R_\alpha R_\alpha^T$ and it is normal. Its characteristic polynomial is

\chi_{R_\alpha}(t)=(t-\cos\alpha)^2+\sin^2\alpha

Its discriminant is

\Delta=4\cos^2\alpha-4=-4\sin^2\alpha<0

A quarter-turn changes every real direction

The blue vector and its orange image are perpendicular. A quarter-turn preserves no direction line of a nonzero real vector.

$x$: $[-1, 1]$
$y$: $[0, 1]$

A Diagonalizable Matrix That Is Not Normal

Consider the following matrix over $\mathbb{R}$ :

A=\begin{pmatrix}1&3\\0&2\end{pmatrix}

Its eigenvalues $1$ and $2$ are distinct, so the matrix is diagonalizable. The corresponding eigenspaces are

E_A(1)=\ker(A-I)=\operatorname{span}\begin{pmatrix}1\\0\end{pmatrix}

Using these eigenvectors as columns gives an explicit diagonalization:

S=\begin{pmatrix}1&3\\0&1\end{pmatrix},\qquad D=\begin{pmatrix}1&0\\0&2\end{pmatrix}

However, the two eigenvectors are not orthogonal because

\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=3\neq0

Direct multiplication also shows that $A$ is not normal:

AA^T=\begin{pmatrix}10&6\\6&4\end{pmatrix}