0
$\begingroup$

I wish to prove the following: if $T:X\to X$ is a nilpotent (and the linear space $X$ is finite-dimensional), then there exists a basis of $X$ such that the matrix representation of $T$ is upper triangular with zero diagonal elements.

I was trying to prove $\{x_1,...,x_{k+1},v, Tv, ..., T^{n-k-2}v\}$ is a basis for $X$ where $x_i$ are basis for $N(T)$ and $v,..., T^{n-k-2}v$ are basis for $R(T)$ where $T^{n-k}=0$ and $n-k$ is minimal. However the matrix representation w.r.t this basis is diagonal. Is there an easy way to prove this ?

I know that for all $T:X\rightarrow X$, there exists a basis that the matrix representation is upper triangular. How is this useful?

$\endgroup$
  • $\begingroup$ You're trying to re-derive the Jordan Canonical Form. If you know this theorem, just apply it to get one or more Jordan blocks with eigenvalue $0$. Each such Jordan block is a representation with respect to a vector $x$ and its non-zero orbits under $A$ given by $x \mapsto^{T} Tx \mapsto^{T} T^{2}x\cdots \mapsto^{T} = 0$. $\endgroup$ – DisintegratingByParts Nov 3 '14 at 6:23
  • $\begingroup$ I know I can use the Jordan form, but how do you show that $J$ has 0 on the diagonal, in other words, zero eigenvalue for nilpotent matrix $\endgroup$ – amathnerd Nov 3 '14 at 6:38
  • $\begingroup$ If $Tx=\lambda x$, then apply $T$ enough times that $T^{n}x=0$ but $T^{n-1}x\ne 0$. Then $\lambda T^{n-1}x=0$ which gives $\lambda = 0$. Alternatively, $p(T)=0$ where $p(\lambda)=\lambda^{n}$ for some positive power $n$. So the minimal polynomial for $T$ divides $\lambda^{n}$, and the only root is $\lambda=0$. $\endgroup$ – DisintegratingByParts Nov 3 '14 at 6:42
2
$\begingroup$

First of all, it is not necessarily the case that the elements $T^j v$ for $j=0,\cdots, n-k-1$ span the range of $T$. You may very well need many $v_i$, but let's just say that one $v$ is enough. Then you need to prove that these elements are linearly independent (where $v$ here is an element such that $T^{n-k-1}v\neq 0$). This is not hard.

Once you do this, you are done, because you have a basis for $X$ (along with the kernel of $T$ as you remark) and all you need to do is form the matrix of $T$ with respect to the basis $\{x_1,\cdots, x_{k+1},v,\cdots T^{n-k-1}v\}$.

For instance, this matrix is (say $k=0$ and $n=3$) $$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$$ just as you wanted.

I think you mistakenly believe that the matrix you are getting is

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}$$ or something similar. This is not correct; if this were true, $T$ would carry $v$ to $v$, $Tv$ to $Tv$ etc. It does not, it shifts the basis elements. Try it yourself with $v=(0,0,1)=e_3$ and $T(e_i)=e_{i-1}$ to get a feel for the form.

In general you need to consider all $v_i$ that form so called cyclic subspaces of $X$ under $T$, take them all together with the kernel and glue the transformations together in a big direct sum. Here is a set of notes I just googled which contains details: http://www.mth.msu.edu/~shapiro/pubvit/Downloads/CycNilp/CycNilp.pdf

$\endgroup$
  • $\begingroup$ I how do you obtain this matrix $\begin{pmatrix} 0 \ 1 \ 0\\ 0 \ 0 \ 1\\ 0 \ 0 \ 0\end{pmatrix}$ above? If $\{x_1, v, Tv\}$ is a basis , so $T^3=0$, we have: $$T(x_1)=0 x_1+0v+0Tv, T(v)=0x_1+ 0v+1Tv, T(Tv)=T^2v\in N(T)$$ so your matrix looks something like this:$\begin{pmatrix} 0 & 0 & X\\ 0 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$. Sorry for any confusion. Your matrix given is what I think the bases should give it to me $\endgroup$ – amathnerd Nov 3 '14 at 15:17
  • $\begingroup$ Jordan decomposition of $A=\left( \begin{array}{ccc} 1 & -\frac{1}{2 x} & 0 \\ x & 0 & y \\ 0 & -\frac{1}{2 y} & -1 \\ \end{array} \right)$ gives $\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0& 0 & 1 \\ 0 &0 & 0 \\ \end{array} \right)$ Notice that $A$ is nilpotent of degree 3 for any non-zero $x$ and $y$. $\endgroup$ – Dr. Wolfgang Hintze Feb 20 at 12:53
0
$\begingroup$

If an upper triangular matrix is nilpotent then its diagonal is zero and hence it is strictly upper triangular. Indeed, if $A$ and $B$ are upper triangular then $C := A\cdot B$ is also upper triangular and the diagonal of $C$ is the component-wise product of the diagonals of $A$ and $B$.

$\endgroup$
  • $\begingroup$ Thanks, but this is not what I am trying to prove $\endgroup$ – amathnerd Nov 3 '14 at 2:58
  • $\begingroup$ @amathnerd: You have an answer for your question: "I know that for all $T:X\to X$, there exists a basis that the matrix representation is upper triangular. How is this useful?"... or just remove it if that is not part of your question anymore ... and I will remove it too, let me know $\endgroup$ – Orest Bucicovschi Nov 3 '14 at 4:30
0
$\begingroup$

The existence of a basis of which $T$ is strictly upper-triangular is equivalent to the existence of a flag $X = X_0 > X_1 > \cdots > X_n = 0$ with $T(X_i) \subset X_{i+1}$. To show the latter, note that any subspace $V$ with $T(V)\subset V$ must have $\ker T\vert V \not =0$ and thus $\text{im}\; T < V$.

$\endgroup$
  • $\begingroup$ What is the $X_i$ I should be looking for $R(T^i)$? $\endgroup$ – amathnerd Nov 3 '14 at 2:55
  • $\begingroup$ Choose the basis $\{e_1, \dots, e_n\}$ so that $X_i = X_{i+1} \oplus \langle{e_i}\rangle$. $\endgroup$ – anomaly Nov 3 '14 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.