I am relatively new to writing proofs and I would appreciate some criticism to improve my proof writing skills. Here is proof to prove that if $A^2 = 0$, then 0 is the only eigenvalue of A for my linear algebra class.
Let $A$ be matrix $M_{nxn}$. If $A^2$ is $M_{nxn}$, then $A^2x = A(Ax)$ according to the associative property of matrix multiplication. $\lambda$ is an eigenvalue of A when there is a nonzero vector x such that $Ax = \lambda x$. Then $A(Ax) = \lambda (\lambda x) \rightarrow A^2x = \lambda ^2 x$. If $A^2 = 0$, then the only value of $\lambda$ must be $0$. $A^2x = \lambda ^2 x \rightarrow 0^2x = \lambda ^2 x$, $\lambda = 0$.
- I am torn on writing "according to the associative property of matrix multiplication". This is very wordy and seems trivial at this point of the class, but I feel like it adds verification to what I am saying and I would rather be safe than sorry.
- It doesn't seem that my proof flows as well from the first half to the second half either. Is there a better way I could be structuring the proof?
You could split up the proof over multiple lines like this:
Let $A \in \mathbb{R}^{n\times n}$ such that $A^2=O$ and let $x$ be an eigenvector of $A$ with eigenvalue $\lambda$. We have: $$\begin{align} & A^2x \\ = \; & A(Ax) \text{ (associative property)} \\ = \; & A(\lambda x) \text{ ($x$ is an eigenvalue of $A$)} \\ = \; & \lambda Ax \\ = \; & \lambda^2 x \text{ ($x$ is an eigenvalue of $A$)} \\ \end{align}$$ Since $A^2x = Ox = 0$, we have $\lambda^2 x = 0$ for the (nonzero) eigenvector $x$, so $\lambda=0$.