2
$\begingroup$

Let $V$ be a 5-dimensional vector space, and $T : V → V$ a nilpotent linear transformation of order $k$, where $1 ≤ k ≤ 5$. Make a list of all the possible dimensions of $\ker(T), \ker(T^2), \ldots , \ker(T^{k−1})$ and the corresponding canonical forms of $T$.

I'm not sure where to start so I started with a few difference branches of thought. Would the eigenvalues here have to be zero because to be nilpotent $T^k=0$? Would the dimensions be as simple as $\ker(T)$ having dimension $1$, $\ker(T^2)$ having dimension $2$, etc?

$\endgroup$
3
  • $\begingroup$ Presumably you know about the Jordan canonical form, based on your question. It is a good idea to start by writing down the possible canonical forms of a nilpotent matrix of order $k$, then compute the kernel for each of these. $\endgroup$
    – Elle Najt
    Mar 22, 2016 at 23:15
  • $\begingroup$ And yes, the eigenvalues are going to be zero. But the answer to the second question is negative in general - I think you are secretly working with a maximal rank nilpotent matrix. $\endgroup$
    – Elle Najt
    Mar 22, 2016 at 23:16
  • $\begingroup$ ker(T) can have any dimension from 1 up to 5. Consider the zero matrix. $\endgroup$ Mar 22, 2016 at 23:34

1 Answer 1

1
+50
$\begingroup$

The dimension of the kernel of $T$ is the number of Jordan blocks in the reduced form of the matrix.

$\dim\ker A^2-\dim\ker A$ is the number of Jordan blocks of size $\ge 2$, &c.

Added:

Jordan's reduction theorem asserts any matrix over an algebraically closed field (say $\mathbf C$) is similar to a block diagonal matrix, where each (Jordan) block is an upper triangular matrix, associated to an eigenvalue $\lambda$ and has the form $$\begin{bmatrix} \lambda&1&0&\dots&0\\0 &\lambda&1&\dots&0\\[-2ex]\vdots&&\ddots&\ddots&\vdots\\0&0&0&\dots&1\\0&0&0&\dots&\lambda \end{bmatrix}$$ For a nilpotent matrix, the only eigenvalue is $0$, and the matrix, in a Jordan basis, has the form $$J=\begin{bmatrix} 0&1&0&\dots&0\\0 &0&1&\dots&0\\[-2ex]\vdots&&\ddots&\ddots&\vdots\\0&0&0&\dots&1\\0&0&0&\dots&0 \end{bmatrix}$$ Note that, $$J^2=\begin{bmatrix} 0&0&1&0&\dots&0\\[-2ex]0 &0&0&1&\ddots&0\\[-2ex]\vdots&&&\ddots&\ddots&\vdots\\0&0&0&0&\dots&1\\0&0&0&0&\dots&0\\0&0&0&0&\dots&0 \end{bmatrix},\quad\&\text c.$$ Hence, if $J$ has dimension $k$, $J^{k-1}\ne 0,\enspace J^k=0$.

$\endgroup$
4
  • $\begingroup$ Would you mind explaining Jordan blocks please? $\endgroup$
    – user319635
    Mar 23, 2016 at 1:39
  • $\begingroup$ @user319635: I've added some details. Is that clearer? $\endgroup$
    – Bernard
    Mar 23, 2016 at 9:41
  • $\begingroup$ Thank you! Okay, so then in my case the dimension of $\ker(T) =1$, the dimension of $\ker(T^2)$, etc, $dim(\ker(T^5))=4$? Is this what they mean by make a list? $\endgroup$
    – user319635
    Mar 24, 2016 at 2:36
  • $\begingroup$ $\dim\ker T=1$ says you have only one Jordan block (necessarily of dimension 5), so $T^5=0$ (and $\dim\ker T^4=5$). If $\dim\ker T$ were equal to $2$, you would have two Jordan blocks, one of dimension $4$ and one of dimension $1$, or one of dimension $3$ and one of dimension $2$. To determine the case, we use the sequence of dimensions. $\endgroup$
    – Bernard
    Mar 24, 2016 at 9:56

Your Answer

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