Problem related to a 2x2 matrix

[boombaby]

Homework Statement

Let A be any real 2x2 matrix. If A^m=0, prove that A^2=0.

Homework Equations

The Attempt at a Solution

It might be quite easy, but I cannot find any way to prove it...all my attempts failed to get any closer to the solution...(I'm not sure why it is the A^2 which should equals to 0, not A^1,or A^3...)

Any hint will be greatly appreciated!

 

[Dick]

Try thinking about what the characteristic polynomial of A might look like. Does A^m=0 tell you anything about possible eigenvalues of A?

 

[boombaby]

this exercise appeared right after the very early chapter where matrix is introduced and eigenvalues are not mentioned...I will review that eigenvalue part though and thanks for the idea .If possible, could you find me an more elementary way to prove it I thought it should have one...
I'm reading <Introduction to Algebra - part I: Fundamentals of Algebra, by A. I. Kostrikin> for a reviewing study.

 

[Dick]

this exercise appeared right after the very early chapter where matrix is introduced and eigenvalues are not mentioned...I will review that eigenvalue part though and thanks for the idea .If possible, could you find me an more elementary way to prove it I thought it should have one...
I'm reading <Introduction to Algebra - part I: Fundamentals of Algebra, by A. I. Kostrikin> for a reviewing study.

Well, if you let V be the space of two dimensional vectors, the A^n(V)=0=A(A^(n-1)(V)). Since A^(n-1)(V) is not equal to zero then there must be a vector v such that A(v)=0. If you write the matrix of A in a basis containing v, then the matrix of A must have a column of zeros. So it must look like [[0,a],[0,b]]. Start writing down powers of A. Do you see that if A^2 isn't zero then no power of A is zero?

 

[boombaby]

I've done it: Let A=(a b; ka kb), assuming rank A=1 (if rank A=0, things are getting trivial, and rand A=2 will never satisfy A^n=0). this yields that A^2=(a+kb)A, which gives A^m=(a+kb)^(m-1) A. Hence if A~=0, a+kb=0, and A^2 therefore equals to 0.
what I also find is that A does not necessarily have a column of zeros, a k satisfying a+kb=0 is the only necessity imo...well, this problem seems really easy, anyway I got a lot more helpful information:) which I'm not quite good at for now, but I will keep an eye on it when I reach those parts. A big thank u that you are always helping me and others :)

 

[Dick]

That works fine. My point was that if B is a matrix representing the same linear transformation as A in a different basis, then A^n=0 etc if and only if B^n=0 etc. So you can do row operations (change of basis) on A. Take your A and multiply the first row by -k and add it to the second. Now you have B=(a b; 0 0). Actually, I like your way better. Less explaining to do.