1. ## Linear algebra

let A be nxn matrix.
Prove:
1. if A^2 = 0 so A columns are vectors in the solution space of the system Ax=0, and that the matrix deg is <= to n/2.
2. if deg(A^2)<deg(A), so to Ax=0 there is non trivial solution and that to A^2x=0 there is y that Ay =/ 0 (Ay is not equal to zero).

2. your question needs to be translated into "mathematics" and "English" first!!

Originally Posted by omert
let A be nxn matrix.
Prove:
1. if A^2 = 0 so A columns are vectors in the solution space of the system Ax=0, and that the matrix deg is <= to n/2.
by "A columns" you mean the "columns space" of A and by "matrix degree" you mean the "rank of A". you also need to use "question mark". i'm not sure if the first part of your question, which is

trivial, is a "question" or you "know" it.

anyway, for the second part, since $A^2=0,$ we have $\text{col}(A) \subseteq \text{nul}(A).$ thus $\text{rank}(A)=\dim \text{col}(A) \leq \text{nullity}(A).$ so by rank-nullity theorem: $n=\text{rank}(A) + \text{nullity}(A) \geq 2 \text{rank}(A).$

2. if deg(A^2) < deg(A), so to Ax=0 there is non trivial solution and that to A^2x=0 there is y that Ay =/ 0 (Ay is not equal to zero).
this time the second part of your question is unclear to me. you need to check it again. for the first part, suppose $Ax=0$ has no non-trivial solution, which means $\text{nullity}(A)=0.$ then obviously

$\text{nullity}(A^2)=0.$ thus $\text{rank}(A)=\text{rank}(A^2)=n,$ which contradicts our assumption that $\text{rank}(A^2) < \text{rank}(A).$

3. sorry for the "bad" english, I'm don't know all the terms in english, so I try to translate(and It's uncorrect sometimes).
I wanted to know how I know if the columns space of A is in the solution space of Ax=0???
also, When you write nul(A) do you mean the solution space of Ax=0???
the second part is:
prove that if: rank(A^2) < rank(A) so to Ax=0 there is non-trivial solution!
and prove that: for A^2x=0 there is y that Ay=/0(non equal to zero)!
I checked!