$A$ is diagonalizable over $F$ if and only if the minimal polynomial of $A$ splits over $F$

Question

I came up with a theorem that

a matrix $A$ is diagonalizable over $F$ if and only if the minimal polynomial of $A$ splits over $F$.

I do not think it is true because we can take the example of the transformation $T(\epsilon_1)=\epsilon_1$ and $T(\epsilon_2)=\epsilon_1+\epsilon_2$ the sheer transformation with minimal polynomial $(t-1)(t-2)$. How can this statement be true then?

Answer 1

The question per se has been answered; this post answers various confusions in the comments.

The OP asks how can a minimal polynomial have repeated roots. "how can this be?" is hard to answer, but in fact a minimal polynomial can have repeated roots. A constructive answer to "how can this be?" might be possible if we knew why someone erroneously thought that a minimal polynomial cannot have repeated roots; then we could explain the error.

I've been asked how one finds the minimal polynomial of a matrix. Never thought about it. Two procedures spring to mind:

(i) Say the characteristic polynomial is $p$. If we can factor $p$ then we can do this: Look at all the monic polynomials $q$ that divide $p$, one by one. For each one, check whether $q(A)=0$. The minimal polynomial is the $q$ of minimal degree such that $q(A)=0$.

(ii), Or, simpler but perhaps less efficient: Let $k$ be the minimal positive integer such that $I,A,\dots,A^k$ are linearly dependent. (To find $k$: test $I,A$ for independence, then test $I,A,A^2$, etc.) A dependence relation $$A^k+c_{k-1}A^{k-1}+\dots+c_0I=0$$gives the minimal polynomial.

Example.

Let $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$.

(i) $p(x)=(1-x)^2$. The only non-trivial monic factor of $p$ is $q(x)=x-1$; since $q(A)\ne0$ the minimal polynomial must be $p$.

(ii) $I,A$ are certainly independent. But $A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$; hence $$A^2+1=2A,$$so the minimal polynomial is $m(x)=x^2-2x+1$.

The Theorem

Having said all that we may as well give a proof of the theorem.

Theorem. Suppose $A$ is an $n\times n$ matrix over the field $K$. TFAE: (i) $A$ is diagonalizable over $K$ (ii) the minimal polynomial $m$ splits over $K$ annd has distinct roots (iii) there exists a polynomial $q$ such that $q$ splits over $K$, $q$ has distinct roots, and $q(A)=0$.

It's trivial that (ii)) implies (iii). And it's very easy to see that (ii) implies (ii), because if $q$ is as in (iii) then the minimal polynomial $m$ must divide $q$.

If $A$ is actually diagonal then (iii) is clear; hence (iii) holds iif $A$$ is just diagonalizable.

Now assume (iii). Say $q(x)=(x-\lambda_1)\dots(x-\lambda_k)$. Since $q(A)=0$ tthe nullspace of $q(A)$ is $K^n$; so the "kernel lemmma" (below) shows that $K^n$ is the direct sum of the nullspaces of $A-\lambda_jI$. That is, $K^n$ is sppannned by eigenvectors of $A$, so $A$ is diagonalizable.

In general let's write $Z(T)$ for the nullspace of $T$.

Lemma. If $p$ and $q$ are relatively prime polynomials then $Z(p(A)q(A))=Z(p(A))\oplus Z(q(A))$.

Proof: Since $K[x]$ is a PID there exist polynomials $r$ and $s$ such that $$pr+qs=1.$$

Suppose $p(A)q(A)x=0$. Now $$x=p(A)r(A)x+q(A)s(A)x,$$and $$q(A)(p(A)r(A)x)=r(A)(p(A)q(A)x)=0.$$Similarly $p(A)(q(A)s(A)x)0,$so $x=p(A)r(A)x+q(A)s(A)x$ shows that $$Z(p(A)q(A))=Z(p(A))+Z(q(A)).$$

Now assume $p(A)x=q(A)x=0$. Then $x=p(A)r(A)x+q(A)s(A)x=0$. So $$Z(p(A))\cap Z(q(A))=\{0\}.$$

Answer 2

The proper statement is that $A$ is diagonalizable iff its minimal polynomial splits and has distinct roots.

In particular the minimal polynomial (thanks to David for pointing out my mistake) of your example is $p(x) = (1 - x)^2$, which does not have distinct roots and so is not a counterexample.