Prove that if $A^2=0$, then $0$ is the only eigenvalue of $A$.

Question

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.
$A^2 = A\times A$ 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 $A$ is 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 $\det(A) = 0$.
$\lambda = 0$,
$\rightarrow$ $\begin{vmatrix} 0I - 0 \end{vmatrix} = 0$

Answer 1

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

If $A$ had an eigenvalue $\lambda\neq0$, then there would be a vector $v\neq0$ such that $A.v=\lambda v$. So,\begin{align}A^2.v&=A.(A.v)\\&=A(\lambda v)\\&=\lambda(A.v)\\&=\lambda^2v\\&\neq0\end{align}and therefore $A^2\neq0$.

Answer 2

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 $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

The rest I cannot even comment on.

What you want to do is the following. Suppose there is an eigenvalue $\lambda\neq 0$. Then there is an eigenvector $v\in\mathbb C^n\setminus\{0\}$ such that $Av=\lambda v$. But then $A^2v=\lambda^2v\neq 0$, so the matrix $A^2$ cannot be $0$.

Answer 3

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 $0 = A^2x = AAx = A\lambda x = \lambda Ax$. 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 $A$.

Answer 4

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 $x^2\implies m_A(x)=x, x^2$. $0$ is the only root of the minimal polynomial, so $A$ just has one distinct eigenvalue: $0$.