📐 Diagonalizability via Minimal Polynomial

$T$ is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors.

Proof: **(⇒)** If $T$ is diagonalizable with eigenvalues $\\lambda_1, \\ldots, \\lambda_k$, then $(T - \\lambda_1 I) \\cdots (T - \\lambda_k I) = 0$ on each eigenbasis vector, hence on all of $V$. The minimal polynomial divides $(x - \\lambda_1) \\cdots (x - \\lambda_k)$.

**(⇐)** If $m(x) = (x - \\lamb...

From: Advanced Linear Algebra

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