3. Eigenstructure and diagonalization

Definition

Eigenvalues and eigenspaces

For $T:V\to V$ over $F$, $\lambda\in F$ is an eigenvalue if $Tv=\lambda v$ for some nonzero $v$. Its eigenspace is $E_\lambda=\ker(T-\lambda I)$. For a matrix $A$, $\lambda$ is an eigenvalue exactly when $\chi_A(\lambda)=0$, where $\chi_A(t)=\det(tI-A)$.

Theorem

Independence and diagonalization

Eigenvectors belonging to distinct eigenvalues are linearly independent. An endomorphism of an $n$-dimensional space is diagonalizable over $F$ if and only if the dimensions of its distinct eigenspaces sum to $n$. For each eigenvalue, geometric multiplicity $\dim E_\lambda$ is at most its algebraic multiplicity as a root of $\chi_A$.

Proof

Why distinct eigenvalues give independence

Suppose a shortest nontrivial relation is $\sum_{i=1}^k c_iv_i=0$, with distinct eigenvalues. Applying $T-\lambda_k I$ gives $\sum_{i=1}^{k-1}c_i(\lambda_i-\lambda_k)v_i=0$, a shorter nontrivial relation, contradiction. Bases of the distinct eigenspaces therefore combine independently. Their union is an eigenbasis exactly when its length is $n$. For the multiplicity bound, extend a basis of $E_\lambda$ to a basis of $V$; the matrix has a $\lambda I$ block in the upper left and a zero block below it, so its characteristic polynomial contains $(t-\lambda)^{\dim E_\lambda}$.

A useful test is: factor $\chi_A$ over the chosen field, find a basis for each eigenspace, and check whether the total dimension is $n$. Equivalently, the polynomial splits and geometric multiplicity equals algebraic multiplicity for every root. A repeated root does not by itself prevent diagonalization, and a polynomial that does not split over $F$ prevents diagonalization over $F$.

Worked example

Repeated eigenvalue with a missing eigenvector

For $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$, $\chi_A(t)=(t-2)^2$. Solving $(A-2I)v=0$ gives $v=(s,0)$, so $\dim E_2=1<2$. Thus $A$ is not diagonalizable. By contrast $2I$ has the same characteristic polynomial but a two-dimensional eigenspace.

If $A=PDP^{-1}$, columns of $P$ are eigenvectors and $A^k=PD^kP^{-1}$ for nonnegative integers $k$. This formula is not a license to diagonalize every matrix. State the field: the real rotation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has eigenvalues $\pm i$ over $\mathbb C$, but no real eigenvalues.

Exercises

1. Diagonalize $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$ and find $A^3$.
2. Decide whether $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is diagonalizable over $\mathbb R$ and over $\mathbb C$.
3. Prove that an $n\times n$ matrix with $n$ distinct eigenvalues in $F$ is diagonalizable over $F$.

Full solutions

1. Eigenvectors are $(1,0)$ for $1$ and $(1,1)$ for $2$. Hence $P=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $D=\operatorname{diag}(1,2)$ and $A=PDP^{-1}$. Therefore $A^3=\begin{pmatrix}1&7\\0&8\end{pmatrix}$.
2. Its characteristic polynomial is $t^2+1$. It has no real root and hence is not diagonalizable over $\mathbb R$. Over $\mathbb C$ it has distinct roots $i,-i$ and is diagonalizable; corresponding eigenvectors are $(1,-i)$ and $(1,i)$.
3. Pick one nonzero eigenvector for each eigenvalue. Distinct-eigenvalue independence gives $n$ independent vectors, hence a basis; in it the matrix is diagonal.

Connection

Inner products add geometry to eigenstructure. Under the spectral theorem, normal or symmetric structure supplies an orthonormal eigenbasis rather than merely an eigenbasis.