$A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this? $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this?
I am not getting enough hints to start.
 A: For any vector space endomorphism $A$, we have $$0=\ker A^0\subseteq \ker A\subseteq \ker A^2\subseteq \ldots\subseteq \ker A^n\subseteq \ker A^{n+1}\subseteq \ldots,$$ simply because $A^nv=0$ implies $A^{n+1}v=AA^nv=0$.
If at any point $\ker A^n=\ker A^{n+1}$, then also $\ker A^{n+1}=\ker A^{n+2}$ etc. because $$v\in\ker A^{n+2}\iff Av\in\ker A^{n+1}\iff Av\in\ker A^n\iff v\in\ker A^{n+1}.$$ In other words, the sequence of kernels of powers of $A$ is eventually stationary and before that it is strictly increasing. It follows that the chain becomes stationary at $A^{\dim V}$ or earlier. For nilpotent $A$ this means that $A^{\dim V}$ must be $0$.
A: If $A$ is nilpotent, then this follows from Cayley-Hamilton: $A^2-\mathrm{tr}(A)A+\det(A)I_2=0$ , and $\mathrm{tr}(A)=\det(A)=0$ since $A$ is nilpotent.
A: Nilpotent implies the trace is $0$.
$$ A^2=\begin{pmatrix} a & b \\ c & -a \end{pmatrix}   \begin{pmatrix} a & b \\ c & -a \end{pmatrix} =  \begin{pmatrix} a^2+bc & 0 \\ 0 & a^2+bc \end{pmatrix}$$
Since that is a multiple of the identity, if that's not $0$ then $A^{2n} \neq 0$ for all $n$. But that's absurd, because $A$ is nilpotent.
A: Since $A$ is nilpotent, both of its eigenvalues are zero, and so (via the existence of the Jordan Normal Form), $A = P^{-1} J P$ for some matrix $P$ and either
$$J = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \qquad \text{or} \qquad J = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$$
In either case, $J^2 = 0$, so
$$A^2 = (P^{-1} J P)(P^{-1} J P) = P^{-1} J^2 P = 0.$$
Remark By essentially the same argument, if $A \in M_n(\mathbb{C})$ is nilpotent, then $A^n = 0$, and this is sharp in the sense that it need not be true that $A^{n - 1} = 0$.
A: Hint: Assume that $A^2\neq 0$ then there exists a smaller integer $n\geq 3$ satisfying $A^n=0$ and $A^{n-1}\neq 0$. Let $x\in \mathbb C^2$ such that $A^{n-1}x\neq 0$.
Noticed then that $(x,Ax,A^2x,\ldots, A^{n-1}x)$ should be a linearly independent family containing $n\geq 3$ elements of $\mathbb C^2$.
A: $A$ is nilpotent so it cannot be injective. This means $dim( Ker A ) \geq 1$ and $Ker A \subset Im A$. We also know that: $dim(Im A) + dim(Ker A) = 2$.
You have two cases:


*

*$dim(Ker A) = 2$  then $A = 0$ and $A^2 = 0$.

*$dim(Ker A) = 1$, then $dim(Im A) = 1$. As both subspaces have same dimension and  $Ker A \subset Im A$ they are equal: $Ker A = Im A$ which leads to $A^2 = 0$.
A: $A$ can be put into upper triangular form, and since $A$ is nilpotent the its eigenvalues are both zero and the upper triangular form will have zeroes on its diagonal. Squaring this matrix gives zero, and so $A^2 = 0$ also.
Note that this same argument gives that any $n$ by $n$ nilpotent matrix over ${\mathbb C}$, when raised to the $n$th power, gives the zero matrix. 
A: If $A^{n}=0$ but $A^{n-1}\ne 0$, then there exists $x \ne 0$ such that $A^{n}x = 0$ but $A^{n-1}x \ne 0$. Then $\{ x, Ax,\cdots, A^{n-1}x \}$ is a linearly independent set. To see why, suppose that
$$
              \alpha_{0}x+\alpha_{1}Ax+\cdots+\alpha_{n-1}A^{n-1}x = 0.
$$
By applying $A^{n-1}$ to the above, we arrive at $\alpha_{0}A^{n-1}x=0$, which gives $\alpha_{0}=0$. Then apply $A^{n-2}$ to conclude that $\alpha_{1}=0$, and so on. Because we're working on a two-dimensional space, then $A^{2} = 0$ must hold in order to avoid a contradiction of dimension.
A: Since $0$ is the only eigenvalue of a nilpotent matrix, its trace (sum of eigenvalues) and determinant (product of eigenvalues) are both $0$. These two relations suggest that $A$ is of the form 
\begin{bmatrix}
a & \frac{-a^2}{c} \\ 
c & -a
\end{bmatrix}
Square it to find that it is the zero matrix.
