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

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.

• T is diagonalizable if and only if it minimal polynomial can be written as $(t-\lambda_1)\cdots(t-\lambda_k)$ where $\lambda_1,\cdots\lambda_k$ are distinct eigenvalues of $T$. 2 days ago
• If zero is not one of the $\lambda_i$ then $T$ is diagonalizable. 2 days ago

## 1 Answer

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 $$\lambda \in \mathbb K$$.