$A^{100} = 0$ implies $A^2 = 0$ when $A$ is $2\times 2$ How to show the following claim
$A^{100} = 0 \implies A^2 = 0$ with $A \in Mat(2 \times 2, K)$

If A is the matrix of a linear map $\phi$ then for all $v \in K^2$ the following identity should be true
$\phi^{99}(v) = A^{99}\cdot v = A^{99} \cdot col_i(A) = 0$
But how to show that $A^2 = 0$?
 A: The $(A^nK^2)_{n\ge0}$ form a sequence of subspaces of $K^2$ which is strictly decreasing until it reaches its minimum.
A: Pick $\lambda$ an eigenvalue of $A$. Then $Ax=\lambda x$ for $x$ a non-zero eigenvector. Inductively you find that $A^{100}x=\lambda^{100}x=0$. Therefore $\lambda^{100}=0$. Hence $A$ has only zero eigenvalues. Since $A$ is a $2\times 2$ matrix, from Cayley Hamilton theorem it follows that $A^2=0$.
A: In general, if a square matrix $A$ satisfies $p(A)=0$ where $p$ is a polynomial, then the minimal polynomial of $A$ divides $p$. Thus, since your $A$ satisfies $p(A)=0$ with $p(x)=x^{100}$, it follows that the minimal polynomial of $A$ divides $x^{100}$. But the degree of the minimal polynomial is $\le$ the size of $A$ which is $2$. Hence, the minimal polynomial is either $x$ or $x^2$, so that $A=0$ or $A^2=0$.
A: The minimal polynomial of $A$ is of degree at most the size $2$ of $A$, and divides $X^{100}$; this means it divides $X^2$, so $A^2=0$.
For an elementary proof that the minimal polynomial $\mu$ of any $n\times n$ matrix $A$ has degree at most $n$ (without using the Cayley-Hamilton theorem) one may argue that for any irreducible factor $P$ of the minimal polynomial the dimension of the kernel of $P(A)$ is at least the degree $d$ of $P$, and so its image has dimension at most $n-d$; induction and the fact that $\mu/P$ is the minimal polynomial of the restriction of $A$ to the image of $P(A)$ complete the proof. But in this concrete situation you don't even need that, as it is clear that the factor $P$ must be $X$, so $P(A)=A$ and $\dim(\ker A)\geq1$ is obvious; therefore it boils down to the argument given by Pierre-Yves Gaillard.
For the record I'll add why $\dim(\ker(P(A)))\geq\deg P$: because $P$ is a nontrivial divisor of $\mu$ it kills some nonzero vector $v$, and then $A^i(v)$ for $0\leq i<\deg P$ are linearly independent (by the irreducibility of $P$) vectors, and they all lie in $\ker(P(A))$ since $A$ commutes with $P(A)$. And that $\mu/P$ is the minimal polynomial of the restriction of $A$ to the image of $P(A)$ is because $(\mu/P)(A)$ must kill that image, and no lower degree polynomial could do that by minimality of $\mu$.
