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Can someone give a proof that all $n\times n$ matrices that are nilpotent of order $n$ are similar?

A matrix $A$ is called nilpotent if there exists some positive integer $k$ such that $A^k$ is the $0$-matrix. The order of nilpotency of a matrix is the smallest $k$ with this property. That is, the question is about matrices such that $A^n=0$ yet $A^{n-1} \neq 0$.

I am new in linear algebra and know up to linear transformation and isomorphisms and matrix representation of transformations. The same question is answered in this site using Jordan Matrix and other things which I do not know. So can anyone give a proof using elementary things which I know.

quid
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Kishalay Sarkar
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  • That can't be true. $0$ is a nilpotent matrix and if $A$ is similar to zero then $A=P0P^{-1} = 0$... – Yanko Aug 30 '19 at 13:57
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    @Yanko Presumably, "order $n$" means that $A^n=0\neq A^{n-1}$. A confirmation would be nice, though. – Arthur Aug 30 '19 at 13:59
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    You won't be able to answer it without developing essentially the Jordan form, or pretending you don't know it and in practice constructing the appropriate basis ad hoc. You're better off waiting until you do have the necessary tools so that you can understand the argument in its proper context. – Arturo Magidin Aug 30 '19 at 14:32
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    I restored the mention of "order" in the post, which was removed by a third-party edit. I also added a clarification, since the terminology seems standard enough to me and it cannot reasonably refer to the dimension of the matrix (another possible meaning) since that's already given. The only other reasonable interpretation I could think of is that OP asked something impossible and added redundant mention of the matrix size that happened to make the question correct, which seems a bit unlikely, and even if so not much is lost. – quid Aug 30 '19 at 15:29

2 Answers2

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If $A$ is an $n\times n$ matrix that is nilpotent of order $n$, then there is a non-zero vector $x$ such that $A^nx =0$ but $A^{n-1}x \ne 0$. So $\{ x,Ax,A^2x,\cdots,A^{n-1}x\}$ is a basis, and the representation of $A$ with respect to this basis has a standard form. This standard form is the same for any such matrix, which makes any two such matrices similar.

Disintegrating By Parts
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First a lemma: Consider the chain $$ \ker(I)\subseteq \ker(A)\subseteq\ker(A^2)\subseteq \cdots\subseteq\ker(A^n) $$ of subspaces of $\Bbb R^n$ (or $\Bbb C^n$ if these are complex matrices). Only the last kernel is the whole space, and the first kernel is the zero space. I claim that each inclusion is strict, and as a consequence the dimension has to increase by exactly $1$ each step.

Proof of claim: Assume that for some $k$ we have $\ker(A^k)=\ker(A^{k+1})$. Let $v\in\ker(A^{k+1})$. Then $Av\in\ker(A^{k+1})=\ker(A^k)$, which is to say $A^k(Ak)=0$, so $v\in\ker(A^{k+1})$. This shows that $\ker(A^{k+1})=\ker(A^{k+2})$, and by induction we see that in the chain above, from $\ker(A^k)$ on, there is only equality. Since $\ker(A^{n-1})\neq\ker(A^n)$, equality can't happen before then.

Corollary: $\ker(A^k)=\operatorname{im}(A^{n-k})$. And in particular, $A$ maps $\ker(A^{k})$ surjectivity onto $\ker(A^{k-1})$.

Now for the actual proof: Let $v_1$ be a non-zero element in the kernel of $A$, and let $v_k$ for $1< k\leq n$ be such that $Av_k=v_{k-1}$. In other words, let $v_2$ be such that $Av_2=v_1$, and so on (by the corollary above, such a vector can always be found). Finally, let $V$ be the matrix where the $k$'th column is $v_{k}$ (so that $v_1$ is the rightmost column). Then $$ V^{-1}AV=\begin{bmatrix} 0&0&0&0&\cdots&0&0\\ 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&1&0\end{bmatrix} $$

Arthur
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  • Why must the rank of $A^k$ decrease at each step, is it possible that for some $k$, $rank(A^k)=rank(A^{k+1})$? – NazimJ Aug 30 '19 at 15:33
  • @NazimJ If that happens, then we also have $rank(A^{k+1})=rank(A^{k+2})$, and so on. Since the rank must reach $0$ at $A^n$, and no earlier, we can't have this happen for $k<n-1$. – Arthur Aug 30 '19 at 15:35
  • Maybe the 'why' of "then we also" is what is unclear. // I added the explanation on the order in OP. If you want you could remove it here. – quid Aug 30 '19 at 15:37
  • @NazimJ if the rank of $A^{k+1}$ is equal to that of $A^k$, then the restriction of $A$ to the image of $A^k$, $A|{\operatorname{im}(A^k)}: \operatorname{im}(A^k) \to \operatorname{im}(A^k)$ is an isomorphism. It follows that for any $n$, $$ A^n|{\operatorname{im}(A^k)} = (A|_{\operatorname{im}(A^k)})^n: \operatorname{im}(A^k) \to \operatorname{im}(A^k) $$ will be an isomorphism and therefore non-zero. – Ben Grossmann Aug 30 '19 at 15:50
  • @NazimJ I reformulated the answer to use intermediate results that are more directly related to the problem at hand. Maybe they are easier to understand too. – Arthur Aug 30 '19 at 16:40