Nilpotent / Diagonalizable matrices

[alaa_137]

Hey guys
I hope I'm in the right place...
I have this question I've been trying to solve for too long:

Let A be an nxn matrix, rankA=1 , and n>1 .
Prove that A is either nilpotent or diagonalizable.

My best attempt was:
if A is not diagonalizable then det(A)=0 then there is a k>0 such that A^k = 0 then A is nilpotent.

But I'm quite sure that's not good...

Anyone can help?

Thanks a lot

 

[micromass]

if A is not diagonalizable then det(A)=0 then there is a k>0 such that A^k = 0 then A is nilpotent.

This is incorrect. This should be clear since you did not use the rank=1 property.

Also, why should a non-diagonalizable matrix have det(A)=0??
And more crucial, why should a matrix with det(A)=0 have A^k=0 for a k?

What does diagonalizable mean?? What does it say about the eigenvectors??

Do you know anything about Jordan canonical forms??

 

[alaa_137]

alright so everything i said is wrong i guess.
i just tried to get to SOMEthing... because i came to a dead end...
and yes i know about Jordan forms...
How can I use that for the proof?

 

[micromass]

Well, try to write down the Jordan canonical forms of a matrix with rank 1. Can you find some eigenvalues of such a matrix?

 

[alaa_137]

if rank = 1 then all eigenvalues are zero, right ?

 

[Ray Vickson]

if rank = 1 then all eigenvalues are zero, right ?

What is the rank of [tex] A = \pmatrix{1&0\\0&0}?[/tex] What are its eigenvalues?

RGV

 

[alaa_137]

t1=1 , t2=0

 

[micromass]

t1=1 , t2=0

So this is a counterexample to your post 5, right?

 

[alaa_137]

yeah, and its jordan form is just as the original matrix.
I tried another rank 1 matrix to see if i come across a rule for jordan forms for rank 1 matrices... but i didn't find anything special...