Finding the Jordan canonical form of a matrix

[nightingale123]

Homework Statement

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##

Homework Equations

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 won't case 2 just be the same as case 1?
Thanks for your help

 

[StoneTemplePython]

Homework Statement

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##

Homework Equations

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 .

View attachment 205270
##\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(P P^{-1})tr(J(A))\\
tr(A)=tr(J(A))\\
##
however the first 10 eigenvalues of J(A) are 0 so won't case 2 just be the same as case 1?
Thanks for your help

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.

The end should read:

##
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##

 

[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.

 

[nightingale123]

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