<s>[INST] Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.
I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.
Thank you very much for your help!
 [/INST] A^k=0 iff A^2=0 k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0 Then A^2=0. Take A^3=A^2A=0. And prove A^2+n+1=0 =>A^2=0 and A^2+nA=0. Then A^2=0.
First direction A^2=0 then A^k=0. 
You can take A and B as a matrix.. Lets represent both matrix . We have A=[a b over c d] and B=[b1 b2 over b3 b4]  then we have B times A=0.
 (multiplying) we have
BA= (b1a+b2c  and  b1b+ b2d  over b3a+b4c and b3b+b4d)
It was the matrix. Then
b1a+b2c=0   b1b+ b2d=0  b3a+b4c=0  b3b+b4d=0
Where we have
(ad-bc)b1=0   (ad-bc)b2=0  (ad-bc)b3=0 and
 (ad-bc)b4=0
Then we have 2 possibilities:
b1=b2=b3=b4=0 or ad-bc=0 this is B=0 or ad=bc
Obviously if A^2=0 then A^k =0 for every k bigger than 2.
A second direction A^k=0 then A^2=0 :
Now let's prove A^k=0  for k>2 with k as positive integer. Then A^2=0.
Watch first k>2, then you can write k=2+n with k positive integer  n>=1.
Let's prove using induction that: 
 if A^2+n=0 then A^2=0
For n=1 we have A^3=A^2.A=0. 
Previously we worked on matrix we can say A^2=0 or ad=bc. 
Let's represent A^3 as matrix:
A^3=[(a^2+bc)a+(ab+bd)c   and  (a^2+bc)b+(ab+bd)d  over (ca+dc)a+(bc+d^2)c  and   (ca+dc)b+(bc+d^2)d)]
It was the matrix. 
A^3=0 then 
(a^2+bc)(a+d)=0    (ab+bc)(a+d)=0
 (ca+dc)(a+d)=0    (bc+d^2)(a+d)=0
Then  let's represent the matrix as : A^2=[a^2+bc  and  ab+bd  over
ca+dc   and  bc+d^2]=[0  and 0  over 0. And 0]
Then 
a=-d but we have ad=bc  then a^2+bc=0 
bc+d^2=0  ab+bd=0  and ca+dc=0 then A^2=0
And prove that A^2+n+1=0 then A^2=0
If A^2+n.A=0 then A^2+n.A=0. Because previous work  A^2+n=0 or ad=bc
 (induction) we conclude A^2+n=0 and 
Because the induction hypothesis we have A^2=0 finally we have A^k=0 then A^2=0. Done
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