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I am proving by construction that there is some basis in which a nilpotent endomorphism has a jordan canonical form that has only ones over supradiagonal. I'll put what I have already and stop where my problem is in order you can think it in the way I am.

What I want to prove is:


Let $T\in\mathcal{L}(V)$ a $r-$nilpotent endomorphism, $V(\mathbb{C})$ a finite-dimensional vector space. There is some basis of $V$ in which the matrix representation of $T$ is a block diagonal matrix, and the blocks have the form \begin{align*} \left( \begin{array}{cccccc} 0 &1 &0 &0 &\dots &0\\ 0 &0 &1 &0 &\dots &0\\ 0 &0 &0 &1 &\dots &0\\ \vdots &\vdots &\vdots &\vdots &\ddots &\vdots\\ 0 &0 &0 &0 &\dots &1\\ 0 &0 &0 &0 &\dots &0 \end{array} \right) \end{align*} that is, blocks that have null entries except for the ones-filled supradiagonal.


First we have that if $T$ is a $r-$nilpotent endomorphism then $T^{r}=0_{\mathcal{L}(V)}$, then, since $U_{1}=T(V)\subseteq V=id(V)=T^{0}(V)=U_{0}$ therefore $U_{2}=T^{2}(V)=T(T(V))\subseteq T(V)=U_{1}$ and if we suppose that $U_{k}=T^{k}(V)\subseteq T^{k-1}(V)=U_{k-1}$ we conclude that $U_{k+1}=T^{k+1}(V)=T(T^{k}(V))\subseteq T(T^{k-1}(V))=T^{k}(V)=U_{k}$. Then we have proven by induction over $k$ that $U_{k}=T^{k}(V)\subseteq T^{k-1}(V)=U_{k-1}$, and since $T^{r}=0_{\mathcal{L}(V)}$, and $U_{k}=T(U_{k-1})$ then $\{0_{V}\}=U_{r}\subseteq U_{r-1}\subseteq\dots\subseteq U_{1}\subseteq U_{0}=V$ and we have shown too that the $U_{k}$ are $T-$invariant spaces and $U_{r-1}\subseteq\ker T$.

In the same manner, let $W_{0}=\ker T^{0}=\ker id=\{0_{V}\}$ and $W_{k}=\ker T^{k}$. Is easy to see that $T(W_{0})=T(\{0_{V}\})=\{0_{V}\}$ therefore $W_{0}\subseteq W_{1}$, moreover $T^{2}(W_{1})=T(T(W_{1}))=T(\{0_{V}\})=\{0_{V}\}$ therefore $W_{1}\subseteq W_{2}$. Then, suppose $W_{k-1}\subseteq W_{k}$, and we see that $T^{k+1}(W_{k})=T(T^{k}(W_{k}))=T(\{0_{V}\})=\{0_{V}\}$ and therefore $W_{k}\subseteq W_{k+1}$ and we conclude we have the chain of nested spaces $\{0_{V}\}=W_{0}\subseteq W_{1}\subseteq\dots\subseteq W_{r-1}\subseteq W_{r}=V$ since $W_{r}=\ker T^{r}=\ker 0_{\mathcal{L}(V)}=V$.

Since we have a chain of nested spaces in which the largest is $V$ itself, then if we choose a basis for the smallest non-trivial (Supposing $U_{r}\neq U_{r-1}$) of them (that is $U_{r-1}$) we can climb chain constructing a basis for the larger spaces completing the basis we have already, what is always possible.

Now, since $U_{r-1}\subseteq\ker T$ then every vector in $U_{r-1}$ is a eigenvector for eigenvalue $0$. Then every basis we choose for $U_{r-1}$ is a basis of eigenvectors. To complete this basis $\{u_{i}^{(r-1)}\}$ to a basis of $U_{r-2}$ (Supposing $U_{r-1}\neq U_{r-2}$) we can remember that $T(U_{r-2})=U_{r-1}$, therefore every vector in $U_{r-1}$ has a preimage in $U_{r-2}$. Then there are some $u_{i}^{(r-2)}\in U_{r-2}$ (maybe many for each $i$ since we don't know $T$ is inyective) such that $T(u_{i}^{(r-2)})=u_{i}^{(r-1)}$. It's to be noted that for fixed $i$ is not possible that $u_{i}^{(r-2)}=u_{i}^{(r-1)}$ since $u_{i}^{(r-1)}$ is an eigenvector associated to eigenvalue $0$ and also every vector in $U_{r-1}$ since they are linear combinations of the basis vectors. Since we have stated they are non unique we can choose one and only one for every $i$. It only remains to see they are linearly independent: let take a linear combination of null vector $\alpha_{i}u_{i}^{(r-1)}+\beta_{i}u_{i}^{(r-2)}=0_{V}$ and let apply $T$ on both sides, $\alpha_{i}T(u_{i}^{(r-1)})+\beta_{i}T(u_{i}^{(r-2)})=\sum_{i}\alpha_{i}0_{V}+\beta_{i}u_{i}^{(r-1)}=\beta_{i}u_{i}^{(r-1)}=0_{V}$. Since the last sum is a null linear combination of linearly independent vectors (since they form a basis for $U_{r-1}$), it implies that $\beta_{i}=0$ for every $i$. Therefore the initial expression takes the form $\alpha_{i}u_{i}^{(r-1)}=0_{V}$ and $\alpha_{i}=0$ for every $i$ by the same argument. We conclude that they are linearly independent.

At this moment we have $\{u_{i}^{(r-1)},u_{i}^{(r-2)}\}$ a linearly independent set of vectors in $U_{r-2}$. If $\dim U_{r-2}=2\dim U_{r-1}$, then we have finished the construction, if not ($\dim U_{r-2}\geq 2\dim U_{r-1}+1$) then we have to choose $u_{j}^{(r-2)}$ with $j=\dim U_{r-1}+1,\dots, \dim U_{r-2}$ that complete the set to a basis of $U_{r-2}$. Again, is in construction of the $u_{i}^{(r-2)}$, we remember that $T(U_{r-2})=U_{r-1}$. Therefore, every vector we choose will have, under $T$, the form $T(v_{j}^{(r-2)})=\mu_{ji}u_{i}^{(r-1)}$. But since we want they to be linearly independent from the $u_{i}^{(r-1)}$ and $u_{i}^{(r-2)}$ we can choose them from $\ker T$, that is we can set $u_{j}^{(r-2)}=v_{j}^{(r-2)}-\mu_{ji}u_{i}^{(r-2)}$ and applying $T$ we obtain $T(u_{i}^{(r-2)})=T(v_{j}^{(r-2)})-\mu_{ji}T(u_{i}^{(r-2)})=\mu_{ji}u_{i}^{(r-1)}-\mu_{ji}u_{i}^{(r-1)}=0_{V}$. Then we only need to see they are linearly independent with the others. Let, again, a null linear combination $\alpha_{i}u_{i}^{(r-1)}+\beta_{i}u_{i}^{(r-2)}+\gamma_{j}u_{j}^{(r-2)}=0_{V}$. First we can apply $T$ both sides: $\alpha_{i}T(u_{i}^{(r-1)})+\beta_{i}T(u_{i}^{(r-2)})+\gamma_{j}T(u_{j}^{(r-2)})=\sum_{i}\alpha_{i}0_{V}+\beta_{i}u_{i}^{(r-1)}+\sum_{j}\gamma_{j}0_{V}=\beta_{i}u_{i}^{(r-1)}=0_{V}$ and therefore $\beta_{i}=0$ for every $i$ since $\{u_{i}^{(r-2)}\}$ is a basis. Then the initial expression takes the form $\alpha_{i}u_{i}^{(r-1)}+\gamma_{j}u_{j}^{(r-2)}=0_{V}$. Note that we have to sets of vectors that are in $\ker T$...

This is the point where I don't see a way to say that the $\alpha_{i},\gamma_{i}=0$ for every $i$ in order to say that they are linearly independent. Any kind of help (hints more than everything else) will be good.

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+1 for the effort! –  Arturo Magidin Jun 26 '12 at 17:01
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1 Answer

up vote 1 down vote accepted

Mostly, you are both on the right track and everything you say is correct, though there are a few spots where a bit more thought could let you be sharper. Let me discuss them first.

You note along the way that "(Supposing $U_r\neq U_{r-1}$)". In fact, we know that for each $i$, $0\leq i\lt r$, $U_{i+1}\neq U_i$. The reason is that if we have $U_{i+1}=U_i$, then that means that $U_{i+2}=T(U_{i+1}) = T(U_i) = U_{i+1}$, and so we have reached a stabilizing point; since we know that the sequence must end with the trivial subspace, that would necessarily imply that $U_i=\{\mathbf{0}\}$. But we are assuming that the degree of nilpotence of $T$ is $r$, so that $U_i\neq\{\mathbf{0}\}$ for any $i\lt r$; hence $U_{i+1}\neq U_i$ is a certainty, not an assumption.

You also comment parenthetically: "(maybe many for each $i$ since we don't know $T$ is injective)". Actually, we know that $T$ is definitely not injective, because $T$ is nilpotent. The only way $T$ could be both nilpotent and injective is if $\mathbf{V}$ is zero dimensional. And since every vector of $U_{r-1}$ is mapped to $0$, it is certainly the case that the restriction of $T$ to $U_i$ is not injective for any $i$, $0\leq i\lt r$.

As to what you are doing: suppose $u_1,\ldots,u_t$ are the basis for $U_{r-1}$, and $v_1,\ldots,v_t$ are vectors in $U_{r-2}$ such that $T(v_i) = u_i$. We want to show that $\{u_1,\ldots,u_t,v_1,\ldots,v_t\}$ is linearly independent; you can do that the way you did before: take a linear combination equal to $\mathbf{0}$, $$\alpha_1u_1+\cdots+\alpha_tu_t + \beta_1v_1+\cdots+\beta_t v_t = \mathbf{0}.$$ Apply $T$ to get $\beta_1u_1+\cdots + \beta_tu_t=\mathbf{0}$ and conclude the $\beta_j$ are zero; and then use the fact that $u_1,\ldots,u_t$ is linearly independent to conclude that $\alpha_1=\cdots=\alpha_t=0$.

Now, this may not be a basis for $U_{r-2}$, since there may be elements of $\mathrm{ker}(T)\cap U_{r-2}$ that are not in $U_{r-1}$.

The key is to choose what is missing so that they are linearly independent from $u_1,\ldots,u_t$. How can we do that? Note that $U_{r-1}\subseteq \mathrm{ker}(T)$, so in fact $U_{r-1}\subseteq \mathrm{ker}(T)\cap U_{r-2}$. So we can complete $\{u_1,\ldots,u_t\}$ to a basis for $\mathrm{ker}(T)\cap U_{r-2}$ with some vectors $z_1,\ldots z_s$.

The question is now is how to show that $\{u_1,\ldots,u_t,v_1,\ldots,v_t,z_1,\ldots,z_s\}$ are linearly independent. The answeer is: the same way. Take a linear combination equal to $0$: $$\alpha_1u_1+\cdots +\alpha_tu_t + \beta_1v_1+\cdots +\beta_tv_t + \gamma_1z_1+\cdots+\gamma_s z_s = \mathbf{0}.$$ Apply $T$ to conclude that the $\beta_i$ are zero; then use the fact that $\{u_1,\ldots,u_t,z_1,\ldots,z_s\}$ is a basis for $\mathrm{ker}(T)\cap U_{r-2}$ to conclude that the $\alpha_i$ and the $\gamma_j$ are all zero as well.

And now you have a basis for $U_{r-2}$. Why? Because by the Rank-Nullity Theorem applied to the restriction of $T$ to $U_{r-2}$, we know that $$\dim(U_{r-2}) = \dim(T(U_{r-2})) + \dim(\mathrm{ker}(T)\cap U_{r-2}).$$ But $T(U_{r-2}) = U_{r-1}$, so $\dim(T(U_{r-2})) = \dim(U_{r-1}) = t$; and $\dim{ker}(T)\cap U_{r-2} = t+s$, since $\{u_1,\ldots,u_t,z_1,\ldots,z_s\}$ is a basis for this subspace. Hence, $\dim(U_{r-2}) = t+t+s=2t+s$, which is exactly the number of linearly independent vectors you have.

You want to use the same idea "one step up": you will have that $u_1,\ldots,u_t,z_1,\ldots,z_s$ is a linearly independent subset of $U_{r-3}\cap\mathrm{ker}(T)$, so you will complete it to a basis of that intersection; after adding preimages to $z_1,\ldots,z_s$ and $v_1,\ldots,v_t$, you will get a "nice" basis for $U_{r-3}$. And so on.

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Thanks and excuse me that I have been absent a lot of time. Your remarks are useful and I'll remember what you said. –  elessartelkontar Jun 29 '12 at 4:23
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