Vrftsjtryk

For any vector space endomorphism $A$, we have $$0=ker A^0subseteq ker Asubseteq ker A^2subseteq ldotssubseteq ker A^nsubseteq ker A^{n+1}subseteq ldots,$$ 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 $$vinker A^{n+2}iff Avinker A^{n+1}iff Avinker A^niff vinker A^{n+1}.$$ 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 becaomes stationary at $A^{dim V}$ or earlier. For nilpotent $A$ this means that $A^{dim V}$ must be $0$.

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.

Nilpotent implies the trace is $0$.

$$ A*A=begin{pmatrix} a & b \ c & -a end{pmatrix} begin{pmatrix} a & b \ c & -a end{pmatrix} = begin{pmatrix} a^2+bc & 0 \ 0 & a^2+bc end{pmatrix}$$

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.

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
$$J = begin{pmatrix}0 & 0 \ 0 & 0end{pmatrix} qquad text{or} qquad J = begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix}.$$
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 $A^{n - 1} = 0$.

$A$ 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 $A^2 = 0$.

Hint: Assume that $A^2neq 0$ then there exists a smaller integer $ngeq 3$ satisfying $A^n=0$ and $A^{n-1}neq 0$. Let $xin mathbb C^2$ such that $A^{n-1}xneq 0$.

Noticed then that $(x,Ax,A^2x,ldots, A^{n-1}x)$ should be a linearly independent family containing $ngeq 3$ elements of $mathbb C^2$.

$A$ 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.

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.

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.