1
$\begingroup$

Let be V a finite dimensional vector space operator and T is a linear operator on V whose characteristic polinomial splits, and let $\lambda_1$, ..., $\lambda_k$ be the distinct eigenvalues of T. Prove that if T is diagonalizable then $rank(T-\lambda_i I_d)=rank((T-\lambda_i I_d)^2)$ for $1\leq i \leq k$.

I thought in construct a basis $ \beta$ of eigenvector of T, such that $[T]_ \beta$ is diagonal to proof that $rank(T-\lambda_i I_d)=rank((T-\lambda_i I_d)^2)$ for $1\leq i \leq k$, but I don't have proved it.

$\endgroup$

1 Answer 1

0
$\begingroup$

It is clear that $\ker(T - \lambda_i \cdot \mathrm{id}) \subseteq \ker(T - \lambda_i \cdot \mathrm{id})^2$. For the other direction, let $v \in V$ and decompose it as $v = v_1 + \ldots + v_k$ where $Tv_j = \lambda_j v_j$. If $v \in \mathrm{ker}(T - \lambda_i \cdot \mathrm{id})^2$ then

$$ (T - \lambda_i \cdot \mathrm{id})^2(v) = (T - \lambda_i \cdot \mathrm{id}) \left( \sum_{j \neq i} (\lambda_j - \lambda_i) v_j \right) = \sum_{j \neq i} (\lambda_j - \lambda_i)^2 v_j = \sum_{\{j \neq i \, | v_j \neq 0\}} (\lambda_j - \lambda_i)^2 v_j = 0.$$

Since the set $\{ v_j \, | \, v_j \neq 0 \}$ is linearly independent (for it consists of eigenvectors of $T$ that correspond to different eigenvalues) and since $\lambda _i \neq \lambda_j \neq 0$ for all $j \neq i$, the set $\{ v_j \, | \, v_j \neq 0 \}$ must be empty, showing that $v = v_i$ and hence $v \in \mathrm{ker}(T - \lambda_i \cdot \mathrm{id})$.

Thus, $\ker(T - \lambda_i \cdot \mathrm{id}) = \ker(T - \lambda_i \cdot \mathrm{id})^2$ and hence

$$\dim \ker(T - \lambda_i \cdot \mathrm{id}) = \dim \ker(T - \lambda_i \cdot \mathrm{id})^2 \implies \mathrm{rank}(T - \lambda_i \cdot \mathrm{id}) = \mathrm{rank} (T - \lambda_i \cdot \mathrm{id})^2. $$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .