Diagonalizability

Definition.

A linear operator $T\in \mathcal{L}(V,V)$ is diagonalizable if its matrix $M(T)$ is diagonal in some basis.

Equivalently, $T$ is diagonalizable if there exist a basis comprised entirely of eigenvectors of $T$; or, $V$ is the direct sum of every eigenspace of $T$; or, the dimension of $V$ is the sum of the dimensions of the eigenspaces.

Theorem.

A linear operator $T$ is diagonalizable if and only if its minimal polynomial $m(x)$ splits into distinct linear factors such that $m(x)=(x-{\lambda }_1)...(x-{\lambda }_k)$ for distinct values ${\lambda }_1,...,{\lambda }_k$.

Proof Sketch.

Suppose $T$ is diagonalizable. Then, there exists a basis $v_1,...,v_n$ of eigenvectors of $T$. Each $v_i$ corresponds to a (not necessarily unique) eigenvalue ${\lambda }_i$. Even if we take just the distinct eigenvalues,

$$(T-{\lambda }_{1}I)...(T-{\lambda }_{k}I)v=0$$

for all $v\in V$ must be true. Since every eigenvalue must be a root of the minimal polynomial, $(x-{\lambda }_1)...(x-{\lambda }_k)$ must be the minimal polynomial.

Conversely, suppose $m(x)=(x-{\lambda }_1)...(x-{\lambda }_k)$. This direction is a little more tricky and requires induction on $dim(V)$. The general idea is to show that $V=null(m(T))$ is equal to a direct sum of eigenspaces. We take $p(x)=x-{\lambda }_1$ and $q=m(x)/p(x)$. Then, we use the fact that $gcd(p(x),q(x))=1$ to show that $null(m(T))=null(p(T))⨁null(q(T))$ (Coprime Polynomials). Lastly, we use our induction hypothesis to show that $null(q(T))$ is also the direct sum of eigenspaces since $q(x)$ is the minimal polynomial of $T$ restricted to $null(q(T))$—a subspace of lower dimension.