1

It's a 4x4 matrix, call it $A$, I could do $A\ne 0$, then $A^2=0$ (in this case) and just be able to stop there. But that involves multiplying matrices!

Now I can get dim(kernel) easily, I just bolt on a column of zeros and row-reduce it, the number of rows of 0 is the dimension of the kernel (in my case $2$)

Kernel of composition of linear transformations now I found this question and I liked it, I especially like the second answer without the tick.

If you consider the 4x4 matrix with 1,1,0,0 on the diagonal (so not quite the identity) - this has a kernel of dimension 2 but it is "stable", obviously applying this again and again will keep the matrix as it is.

So in that case $im(A)\cap\ker(A)$ is empty (using their conclusion) which is how dim(kernel($A^2$)) = dim(kernel($A$))

This means in my case that the image and kernel are not disjoin, infact they overlap (with dimension 2).

I'd like to be able to prove when this happens and know more about it. With the almost-identity example something trapped in the (x,y) plane cannot leave it, it is "stable" there, with my matrix that is not the case. because $A^2=0$ I know that apply it again and no matter what vector goes in I will get 0.

What is going on here?

Community
  • 1
Alec Teal
  • 5,590
  • use \cap for intersection – Ellya Apr 22 '14 at 17:19
  • 2
    I don't understand the question. You have a 4x4 matrix, and what exactly is it you know about it and what is it you want to prove about it? – G. Bach Apr 22 '14 at 17:21
  • You should look up the Jordan Normal Form of a matrix. What you're asking about is basically what are the block sizes, or equivalently the dimensions of the generalized eigenspaces of the matrix. – Jim Apr 22 '14 at 17:21
  • @Jim that's actually what I'm working out, I'm going to have to do this multiplication .... I can feel it. – Alec Teal Apr 22 '14 at 17:23
  • In general, for an $n \times n$ matrix, the kernel will stabilize by the $n^\text{th}$ power, but could potentially stabilize at any number before that. There are some tricks involving the characteristic polynomial and minimal polynomials that let you narrow down the choices, but I know of no algorithm for figuring out when the kernel stabilizes that works in general and doesn't involve computing powers of the matrix. – Jim Apr 22 '14 at 17:25
  • @Jim if I have characteristic polynomial $(\lambda-1)^4$ for a 4x4 matrix. Can the minimum polynomial be 3? 3 doesn't divide 4 you see (but I can't find one to test this!) Also if my minimum is of order 2 does that mean I have 2 Jordan blocks (each being 2x2 with 1s on the diagonal and 1 in the upper right?) So if I had $(A-I)=0$, then it'd be 4 blocks and just the identity matrix? – Alec Teal Apr 22 '14 at 17:30
  • If your characteristic polynomial is $(\lambda - 1)^4$ then your minimal polynomial can be $(\lambda - 1)^i$ for $i = 1, 2, 3, 4$. You can basically get any partition, so your blocks could be size $4$, $(3, 1)$, $(2, 2)$, $(2, 1, 1)$ or $(1, 1, 1, 1)$. If your minimum polynomial is order two it means none of your blocks are of size greater than two, but that still leaves $(2, 2)$, $(2, 1, 1)$ or $(1, 1, 1, 1)$ (though that last one is ruled out if there is only one eigenvalue, as your characteristic polynomial indicates). – Jim Apr 22 '14 at 18:12

1 Answers1

0

If $A^2=0$ then you have a nilpotent matrix of index 2, and then the only eigenvalue of $A$ is $0$. Since $A\neq 0$ we also then have that the minimal polynomial can't be $m_A(x)=x$, and so in fact the minimal polynomial is $m_A(x)=x^2$. So then $A$ has one of two possible Jordan canonical forms:

Dg$\left(\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}\right)$ or Dg$\left(\begin{bmatrix}0&0\\1&0\end{bmatrix},0,0\right)$.

If you say that the characteristic polynomial is $(\lambda-1)^4$ then $A^2$ cannot be $0$.

Christiaan Hattingh
  • 4,855