Let A be a 2x2 matrix.
Then, prove the following:
$\displaystyle A^n=0 \Longrightarrow A^2=0$
Hint: $\displaystyle p(x)=x^n$ is an annihilator polynomial of $\displaystyle A$ , this means that the minimal polynomial of $\displaystyle A$ is $\displaystyle \mu(x)=x$ or $\displaystyle \mu(x)=x^2$ so the canonical form of $\displaystyle A$ is $\displaystyle J=\begin{bmatrix}{0}&{0}\\{0}&{0} \end{bmatrix}$ or $\displaystyle J=\begin{bmatrix}{0}&{1}\\{0}&{0} \end{bmatrix}$
This might work, but check my computations. Since $\displaystyle A^n=0$, $\displaystyle \mbox{det}A^n=0\Rightarrow \mbox{ det}A=0$, so if we call the entries in A a,b,c,d, we get ad=bc. Now this is the part you need to check- it looks like using this gives a nice expression for any power of A, with entries involving powers of (a+d). I think this should do it.
In a more auto content way we can reason in the following way: if $\displaystyle A^n=0$ then $\displaystyle \det A=0$ . This means that $\displaystyle A=0$ (and the result is trivial) or $\displaystyle \textrm{rank}A=1$ i.e. $\displaystyle A=\begin{bmatrix}{a}&{\lambda a}\\{b}&{\lambda b} \end{bmatrix}=\begin{bmatrix}{a}\\{b} \end{bmatrix}\begin{bmatrix}{1}&{\lambda} \end{bmatrix}$ with $\displaystyle \begin{bmatrix}{a}&{b} \end{bmatrix}\neq \begin{bmatrix}{0}&{0} \end{bmatrix}$ . So:
$\displaystyle A^n=\ldots=(a+\lambda b)^{n-1}A=0\Leftrightarrow (A=0) \vee (a=-\lambda b)$
If $\displaystyle A=0$ , then $\displaystyle A^2=0$ . If $\displaystyle a=-\lambda b$ then, $\displaystyle A=\begin{bmatrix}{-\lambda b}&{-\lambda^2b}\\{b}&{\lambda b} \end{bmatrix}$ and also $\displaystyle A^2=0$ .
This can be done with quite simple linear algebra.
Since $\displaystyle A^n = 0$, there is a minimal $\displaystyle m$ such that $\displaystyle A^m=0$.
$\displaystyle A^{m-1} \ne 0$ and therefore there is a vector $\displaystyle v$ for which $\displaystyle A^{m-1}v \ne 0$.
It's easy to prove that the set $\displaystyle \{v,Av,A^2v,\cdots,A^{m-1}v\}$
is linearly independent, and contains $\displaystyle m$ elements.
Can you finish now?
In my previous post we can avoid the concept of rank : if $\displaystyle A=\begin{bmatrix}{a}&{c}\\{b}&{d} \end{bmatrix}$ with $\displaystyle \det A=0$ and $\displaystyle (a,b)\neq (0,0)$ then, $\displaystyle (c,d)=\lambda (a,b)$ etc. So, the problem can be solved using only the concepts of determinant and product of matrices, obtaining more simple algebra . I don't mean exactly linear algebra (some say Linear Algebra ends when Cramer's Rule starts) .
Isn't it "where determinants start"? I read an interesting paper lately: http://www.axler.net/DwD.pdf It was a revelation to me, a mediocre student who instantly falls asleep on seeing formulas with many indices.
As for the problem, do you think it's tractable by means of finding a formula for $\displaystyle A^n$ dependent on a, b, c and d? I've just tried it and I can't but I'm not very good.
Well, that depends on the way we categorize.
The characteristic polynomial for a $\displaystyle 2\times 2$ matrix is $\displaystyle \chi (\lambda)=\lambda ^2-\textrm{tr}(A)\lambda +\det A$ . If we know that $\displaystyle \det A=0$ then, $\displaystyle \chi (\lambda)=\lambda(\lambda-\textrm{tr}A)$ . Now, we can use the Cayley-Hamilton theorem. For example for $\displaystyle \textrm{tr}A\neq 0$ you'll obtain $\displaystyle A^n=(\textrm{tr}A)^nA$ .As for the problem, do you think it's tractable by means of finding a formula for $\displaystyle A^n$ dependent on a, b, c and d? I've just tried it and I can't but I'm not very good.