Krdytkyu

Does my proof hold up to prove that $0$ is the only eigenvalue of $A$ if $A^2 = 0$?

Let $A$ be an $n times n$ matrix.
$A^2 = A*A$ because of matrix multiplication.

If $A = k$, where $k neq 0$, then $A^2 neq 0$. Therefore if $A^2 = 0$, $A$ is the zero matrix.

If $A$ = the zero matrix, det($A^2) = 0$.
$lambda neq 0$ if A is non-singular. A is a non-singular matrix IFF det(A) $neq 0$. The charcteristic equation is false when $lambda = 0$ for a non-singular matrix so it can only be true when the matrix is singular, making $det(A) = 0$.
$lambda = 0$,
$rightarrow$
$begin{vmatrix}
0I - 0
end{vmatrix}
= 0$

It is wrong. You seem to believe that if $Aneq0$, then $A^2neq0$, which is false. Take $A=left(begin{smallmatrix}0&1\0&0end{smallmatrix}right)$, for instance.

If $A$ had an eigenvalue $lambdaneq0$, then there would be a vector $vneq0$ such that $A.v=lambda v$. So,$$A^2.v=A.(A.v)=A(lambda v)=lambda(A.v)=lambda^2vneq0$$and therefore $A^2neq0$.

What you wrote is a mess.

If $A = k$, where $k neq 0$, then $A^2 neq 0$. Therefore if $A^2 = 0$, $A$ is the zero matrix.

This is false: consider $A=begin{pmatrix}0&1\0&0end{pmatrix}$.

The rest I cannot even comment on.

What you want to do is the following. Suppose there is an eigenvalue $lambdaneq 0$. Then there is an eigenvector $vinmathbb C^nsetminus{0}$ such that $Av=lambda v$. But then $A^2v=lambda^2vneq 0$, so the matrix $A^2$ cannot be $0$.

There are non-zero matrices $A$ where $A^2 = 0$ so your proof does not hold true. If $A^2 = 0$, we already know that $A$ is non-invertible because if $A$ were invertible, we could multiply both sides with $A^{-1}$ and get $A = 0$, but $0$ is not invertible. We do know $0$ is an eigenvalue of $A$ though because we can multiply $A$ with any column of $A$ to obtain $0$. We now have to show $0$ is the only eigenvalue. Suppose there is an eigenvalue $lambda neq 0$. Then, $Ax = lambda x$ and $0 = A^2x = AAx = Alambda x = lambda Ax$. Since $lambda neq 0$, we obtain $Ax = 0 = 0x$ for some eigenvector $x$, but this means $0$ is an eigenvalue for $x$, which contradicts $lambda neq 0$. Thus, $lambda = 0$ is the only eigenvalue for $A$.

Here is an alternate proof: $x^2$ is an annihilating polynomial for the matrix $A$, so the minimal polynomial for $A, m_A(x)$ divides $x^2implies m_A(x)=x, x^2$. $0$ is the only root of the minimal polynomial, so $A$ just has one distinct eigenvalue: $0$.