$u$ is diagonalizable if and only if its minimal polynomial is split over $K$ and every subspace stable under $u$ has a stable complement.

Question

I would like to demonstrate the following result:

Let $K$ be a commutative field, $E$ a finite-dimensional $K$-vector space. Consider $u \in \mathcal L(E)$. Show that $u$ is diagonalizable if and only if its minimal polynomial is split over $K$ and every subspace stable under $u$ has a stable complement.

If $u$ is diagonalizable, proving the result is not difficult.

I find the question more challenging when proving the converse. I am inclined to use induction, where the induction hypothesis is $\mathcal H(k)$: "there exists a linearly independent family $(e_1, ..., e_k)$ consisting of eigenvectors of $u$." I am struggling to prove $\mathcal H(1)$ because I cannot find a hyperplane stable under $u$. (I am aware that characteristic subspaces are stable since the annihilating polynomial is split...)

Answer 1

Let $\lambda$ be an eigenvalue of $u$; it exists since the characteristic polynomial splits. Write the characteristic polynomial as $u(t) = (t-\lambda)^aq(t)$, where $q(\lambda)\neq 0$ and $a\gt 0$. Let $V$ be the eigenspace of $\lambda$. We want to show that $\dim(V)=a$.

Since $V$ is stable, it has a $u$-stable complement $W$. Because $W$ is $u$-stable, it is also $(u-\lambda I)$-stable. And because $W\cap V=\{\mathbf{0}\}$, then $u-\lambda I$ is one-to-one on $W$.

Now let $\beta$ be a basis for $V$, and let $\gamma$ be a basis for $W$. Considering the coordinate matrix of $u$ relative to $\beta\cup\gamma$, we see that the characteristic polynomial of $u$ has the form $(t-\lambda)^{\dim(V)}r(t)$, where $r(t)$ is the characteristic polynomial of $u|_W$. In particular, $r(\lambda)\neq 0$. By unique factorization, from $$(t-\lambda)^aq(t) = (t-\lambda)^{\dim(V)}r(t),\qquad q(\lambda)\neq 0, r(\lambda)\neq 0,$$ it follows that $q(t)=r(t)$ and $a=\dim(V)$. The latter is what we needed to show.

Since the characteristic polynomial splits, this shows that every eigenvalue has geometric multiplicity equal to its algebraic multiplicity, and the geometric multiplicities add up to $\dim(E)$. Therefore, $u$ is diagonalizable.

Note that I used the splitting of the characteristic polynomial rather than the minimal, but the two are equivalent since any irreducible factor of the characteristic polynomial must divide the minimal polynomial.

Answer 2

Hint : induction on the degree of the minimal polynomial.

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$$\mathcal H(k): \forall E,\, \forall u\in \mathcal L(E),\, \forall \lambda_1,\ldots,\lambda_k\in K,\,\text{if the minimal polynomial of $u$ is}\,\prod_{i=1}^{k} (X-\lambda_i)\,\text{and}\, u\,\text{satisfy the stability property then $u$ is diagonalizable}.$$