6. Spectral theorems

All spaces are finite-dimensional over $\mathbb R$ or $\mathbb C$.

The complex existence argument uses the fundamental theorem of algebra: every nonconstant complex polynomial has a complex root. That separate prerequisite is not proved in this chapter. In dimension zero the empty basis proves the conclusion; the eigenvector step below is used only in positive dimension.

If $S$ is normal, then $\|Sv\|=\|S^\dagger v\|$, because the squares are determined by $S^\dagger S=SS^\dagger$. For normal $T$, the operator $S=T-\lambda I$ is also normal. Thus

$$Tv=\lambda v\quad\Longrightarrow\quad T^\dagger v=\bar\lambda v.$$

If $v$ is an eigenvector, its orthogonal complement is invariant under both $T$ and $T^\dagger$ by the adjoint identity. Their restrictions to that complement remain adjoints and commute, hence are normal.

Complex spectral theorem. A complex endomorphism is normal exactly when it admits an orthonormal eigenbasis.

Proof. In nonzero dimension the characteristic polynomial has a complex root, so choose and normalize an eigenvector. The preceding orthogonal-complement argument reduces to a normal operator on a space of one less dimension. Induction gives the orthonormal eigenbasis. Conversely a diagonal matrix in an orthonormal basis commutes with its conjugate transpose, so is normal. ∎

A self-adjoint eigenvalue is real: $\langle Tv,v\rangle=\langle v,Tv\rangle$ gives $\lambda\|v\|^2=\bar\lambda\|v\|^2$. A unitary eigenvalue has modulus one because norms are preserved. These are necessary spectral properties, not sufficient conditions for an arbitrary matrix to be self-adjoint or unitary.

Real symmetric spectral theorem. A real self-adjoint map has an orthonormal real eigenbasis. For existence of a real eigenvalue, complexify its real symmetric matrix. Any complex eigenvalue is real by the previous calculation. For a complex eigenvector, at least one of its real and imaginary parts is a nonzero real eigenvector for that real eigenvalue. Normalize it and induct on its invariant real orthogonal complement. A real normal operator need not have real eigenvalues: a plane rotation is the standard counterexample.

The real normal condition alone is not enough for a real eigenbasis: a nontrivial plane rotation is real orthogonal and normal but may have no real eigenvectors. Over $\mathbb C$, a normal operator's eigenvalues need not be real; self-adjointness is the additional condition that makes them real.

Here the shared notation $T^\dagger$ is the same Hermitian adjoint as $T^*$ in the preceding unit. An orthonormal eigenbasis has a unitary column matrix $U$, so its diagonalization is $A=UDU^*$. In the real case it is $A=QDQ^{\mathsf T}$ with $Q$ orthogonal and $D$ real diagonal.

Worked example

Orthogonal diagonalization

Let $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$. The orthonormal eigenvectors $q_1=(1,1)/\sqrt2$ and $q_2=(1,-1)/\sqrt2$ have eigenvalues $3$ and $1$. Thus, with $Q=(q_1\ q_2)$, $A=Q\operatorname{diag}(3,1)Q^{\mathsf T}$. For example, $A^k=Q\operatorname{diag}(3^k,1)Q^{\mathsf T}$ for every nonnegative integer $k$.

Choosing a method and avoiding mistakes

First check the field and structure. For real symmetric matrices, find eigenspaces, choose orthonormal bases within each repeated-eigenvalue eigenspace, then combine them; different eigenspaces are already orthogonal. A merely diagonalizable matrix need not admit an orthonormal eigenbasis. $Q^{-1}=Q^{\mathsf T}$ is valid only for orthogonal $Q$, and in complex spaces the corresponding identity uses $U^*$.

Exercises

1. Orthogonally diagonalize $\begin{pmatrix}4&0\\0&-1\end{pmatrix}$ and compute its square.
2. For the matrix in the worked example, find the maximum of $x^{\mathsf T}Ax$ over real unit vectors $x$ and all maximizers.
3. Give an example of a real normal matrix that is not diagonalizable over $\mathbb R$ and explain why this does not contradict the complex theorem.

Full solutions

1. It is already diagonal: choose $Q=I$ and $D=\operatorname{diag}(4,-1)$. Its square is $\operatorname{diag}(16,1)$.
2. Write $x=a q_1+b q_2$ with $a^2+b^2=1$. Then $x^{\mathsf T}Ax=3a^2+b^2=1+2a^2\le3$. Equality holds exactly when $b=0$, so the maximizers are $\pm q_1$.
3. The quarter-turn matrix $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is orthogonal and therefore normal, but its characteristic polynomial $t^2+1$ has no real roots. Over $\mathbb C$ it does have an orthonormal eigenbasis, as the complex theorem asserts.

Connection

The same orthogonal change of variables classifies real quadratic forms. That topic and Jordan form are optional supplements here, not a claim about any particular current examination.