Nilpotent matrix wit h index 2

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Homework Statement


Determine a and b such that A is nipotent of index 2.

A:= <<a,b>|<0,0>>

A is a 2x2 matrix column 1 is a and b , column 2 is 0's


Homework Equations


A^k=0 to be nilpotent and to be nipotent it has to be that A^(k-1) doesn't equal 0..


The Attempt at a Solution


so far I've tried squaring the matrix and getting a result matrix

squared:= <<a^2,ba+b^2>|<0,0>>

so i can only assume that a and b must be zero to insure A^k (in this case k is 2 so squared) but then on A^(2-1) A must not be a zero matrix.. so...

my class didn't cover nilpotent yet and i guess the teacher just assumed we knew it.. so after doing research that's what i have gotten to understand about it.. but would like some help if someone can point out what I'm missing to make it nilpotent and work on both checks the A^k=0 and A^(k-1) not equal zero

Yes it is a homework problem, but i did make an attempt at it for 3hrs before asking for help on it, and its not due for a couple days so its not like I'm just saying here solve it for me.. I'm just looking for some help to understand inpontents cause there are some other problems, like 3 on the matter and i just don't get it i guess.
 
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Unless I'm missing something huge here, you're overcomplicating things. You have a given index (and a "small" one!), and you only have to work one simple matrix equation out (eg matrix is A nilpotent of index 2 if A^2 = 0).
 
but to get it to be nilpotent, a^2=0 and a^(2-1)doesn't eqaul 0, a and b must be given values.. but the only values I've found to work to solve the first part.. proving the second part of the nipotent seems impossible for 2.. because so far a and b being =0 kills the second part of the check
 
The square of your matrix is <<a^2,b*a>|<0,0>>. Now does b need to be zero?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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