Prove or disprove if $m(x)=\prod_{j=1}^{s}(x-\lambda_{j}^{2})$ is the minimal polynomial of a linear map $T\circ T:V\to V$

Question

Prove or disprove:

If $m(x)=\prod_{j=1}^{s}(x-\lambda_{j}^{2})$ (with $\lambda_i\not=\lambda_j$ for $i\not=j$) is the minimal polynomial of a linear transformation $T\circ T:V\to V$, with $V$ a $\mathbb{K}$-vector space of dimension $n$, and $\lambda_j\in\mathbb{K}$ $\forall j=1,\ldots,s$, then $T$ is diagonalizable.

My attempt:

Since $m(x)$ is the minimal polynomial of $T^{2}=T\circ T$, then by Cayley-Hamilton theorem: $$m([T]^{2})=\prod_{j=1}^{s}([T]^{2}-\lambda_{j}^{2}I^{2})=0$$ Where $I$ is the identity matrix and $[T]$ the matrix associated to $T$. So: $$\prod_{j=1}^{s}([T]+\lambda_{j}I)\cdot\prod_{j=1}^{s}([T]-\lambda_{j}I)=0$$ Then $T$ is diagonalizable iff $2s\leq n$.

I think my conclusion isn't correct, however I can't think of another way to develop the problem. I would greatly appreciate your help.

Answer 1

Note: we also suppose that $\lambda_i^2 \neq \lambda_j^2$ for $i \neq j$ below.

In a finite dimensional space $V$ of dimension $n$, a linear transformation is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors.

Considering what you noticed in the question, the minimal polynomial $\mu(x)$ of $T$ divides $M(x)=m(x^2)$. If $0$ is not one of the $\lambda_i$ and the characteristic of $\mathbb K$ is not equal to $2$, then $M$ splits in distinct linear factors (as $-\lambda_i \neq \lambda_i$), hence $\mu$ also and $T$ is diagonalizable.

If zero is one of the $\lambda_i$, then it exists $v \neq 0$ such that $T^2(v)=0 $ and $T(v)\neq 0$ and $T$ is not diagonalizable.

If the characteristic of $\mathbb K$ is equal to $2$, then $T$ is never diagonalizable for a similar reason as above paragraph as in that case $$x^2-\lambda^2=(x-\lambda)^2$$ for any $\lambda \in \mathbb K$.