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Prove any $n \times n$ matrix can be written as in block form:

$\begin{pmatrix} N & 0 \\ 0 & B \end{pmatrix}$

where $N$ is a $k \times k$ nilpotent matrix ($N^n=0$) and $B$ is an $(n-k)\times(n-k)$ invertible matrix.

Here's a link to someone else's solution to the problem:

Matrix similar to block diagonal matrix with a nilpotent and an invertible block

My question is, is there any more elementary approach without using Jordan decomposition? (For instance, using the properties of $null(T^k)$ and $Im(T^k)$?)

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Since $T$ maps each $x$ to $0$ or nonzero, $\dim(T^{k+1})\leqslant\dim(T^{k})$. If inequalities hold for all $n$, i.e, $$ \dim(T^{})>\dim(T^{2})>\cdots>\dim(T^{n}) $$ Then $\dim(T^{n})=0$, which means matrix is nilpotent.

If for some $k$ such that $\dim(T^{k+1})=\dim(T^{k})$, then $T^{k}$ is invariant under $T$. So there is $$ \operatorname{Im}(T^{k})\cap \operatorname{Ker}(T^{k})=\{0\} $$ for if not, let $x\in \operatorname{Im}(T^{k})\cap \operatorname{Ker}(T^{k})$, then $T(x)=0$ and $\dim(T^{k+1})<\dim(T^{k})$.

Since $\dim(\operatorname{Im}(T^{k}))+\dim(\operatorname{Ker}(T^{k}))=n$, plus $\operatorname{Im}(T^{k})$ and $\operatorname{Ker}(T^{k})$ are invariant under $T$, $T$ can be decomposed into $B$ (bases of $\operatorname{Im}(T^{k})$) and $N$ (base of $\operatorname{Ker}(T^{k})$). Clearly $\operatorname{Rank}(B)=\dim(\operatorname{Im}(T^{k}))$ and $N$ is nilpotent.

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