Finding the Jordan canonical form of a matrix

Tags:
1. Jun 11, 2017

nightingale123

1. The problem statement, all variables and given/known data
About an endomorphism $A$ over $\mathbb{C^{11}}$ the next things are know.
$$dim\, ker\,A^{3}=10,\quad dim\, kerA^{2}=7$$
Find the
a) Jordan canonical form of $A$
b) characteristic polynomial
c) minimal polynomial
d) $dim\,kerA$
When:
case 1: we know that $A$ is nilpotent
case 2: we know that $tr(A)=0$
2. Relevant equations

3. The attempt at a solution
So case 1:
$A$ is nilpotent
therefore we know that there exists some number $n$ such that $A^{n}=0$ and since $A^{3 }\neq0$ that must mean that $A^{4}$ must equal 0.
( n cannot be greater than 4 because then the dimension of its kernel would exceed the dimension of $\mathbb{C^{11}}$
so taking into account $dim\, ker\,A^{3}=10,\quad dim\, kerA^{2}=7$ I get .

$\begin{bmatrix} 0&1&&&&&&&&&\\ &0&1&&&&&&&&\\ &&0&1&&&&&&&\\ &&&0&&&&&&&\\ &&&&0&1&&&&&\\ &&&&&0&1&&&&\\ &&&&&&0&&&&\\ &&&&&&&0&1&&\\ &&&&&&&&0&1&\\ &&&&&&&&&0&\\ &&&&&&&&&&0\\ \end{bmatrix}$
characteristic polynomial $p(\lambda)=\lambda^{11}$
minimal polynomial $m(\lambda)=\lambda^{4}$
$dim\,kerA=4$

Case 2:
$tr(A)=0$ here is where I get confused.
I know that
$A=PJ(A)P^{-1}$
therefore
$tr(A)=tr(PJ(A)P^{-1})\\ tr(A)=tr(P)tr(J(A))tr(P^{-1})\\ tr(A)=tr(PP^{-1})tr(J(A))\\ tr(A)=tr(J(A))\\$
however the first 10 eigenvalues of J(A) are 0 so wont case 2 just be the same as case 1?

2. Jun 11, 2017

StoneTemplePython

a few things. One is I'm a bit concerned that you're treating the trace operation like a determinant... They are related and important, but rather different. Trace does have a cyclic property, but in general does *not* allow you to split it apart and multiply like determinants.

$tr(A)=tr(PJ(A)P^{-1})\\ tr(A)=tr(P^{-1}PJ(A))\\ tr(A)=tr(I J(A))\\ tr(A)=tr(J(A))\\$

I may have mis-understood your question about part 2, but I'll give it a go anyway.

Here is the difference between a nilpotent matrix and case 2. For n x n matrices:

Case one:

$trace\big( (A^k)\big) = 0$

for $\{k = 1, 2, 3, ... , n\}$

(you can actually keep counting out all natural numbers but I am happy to stop on k = n).

At some point it is worth proving that that an n x n matrix is nilpotent if and only if

$trace\big( (A^k)\big) = 0$
for $\{k = 1, 2, 3, ... , n\}$

Case two:

$trace\big( (A^1)\big) = 0$

$trace\big( (A^r)\big) = ?$
for $\{r = 2, 3, ... , n\}$

simple example for case 2
$A = \begin{bmatrix} 1 & 2\\ 0 & -1 \end{bmatrix}$

This is not nilpotent. But $trace\big( (A^1)\big) = 0$

Last edited: Jun 12, 2017
3. Jun 12, 2017

StoneTemplePython

follow-up (edit button is disabled)

I see what you are saying now.

$A^3$ has 10 linearly independent vectors in its nullspace, and one vector not in the nullspace.

hence $trace\big(A^3\big) = \lambda_1^3$

But we know $trace\big(A \big) = \lambda_1 + \big(\sum_{k=2}^n \lambda_k\big) = 0$

but each $\lambda_k$ must be zero or else $A^3$ cannot be a rank one matrix. (When you look at $A$ as being similar to an upper triangular matrix, whether Jordan Form or Schur decomposition, you can lower bound its rank by the number of non-zero eigenvalues -- a basic Gaussian Elimination and pivot counting argument --, so if there were 2 non-zero eigenvalues it would be at least rank 2, but we know $A^3$ is rank one.)

$trace\big(A \big) = \lambda_1 + \big(0\big) = 0$, thus $\lambda_1 = 0$

and we have $\lambda_1 = 0$, which brings back the nilpotent case. A bit of a trick question I suppose.

Last edited: Jun 12, 2017
4. Jun 12, 2017

nightingale123

thank you for point out the trace mistake I completely missed that