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A problem I've encountered is this.

Prove that for any standard matrix $A$ of any linear transformation, if $A^2=0$, the only eigenvalue of $A$ is zero.

By definition, if $\lambda$ is an eigenvalue of $A$, then $Ax=\lambda x$ for some nonzero eigenvector $x$.(We don't have to really worry about the zero eigenvector, since that is a trivial eigenvector for any linear transformation.) So $A^2 x=A(\lambda x)=0x=0$. From that we can pull out $A(\lambda x)=0$.

Now assume that there is some nonzero $\lambda$ such that $A(\lambda x)=0$. Since $x$ is an eigenvector of $A$, any vector in $Span(x)$ is also an eigenvector of $A$. Since $\lambda$ is an eigenvalue, it is a scalar, and thus $\lambda x$ is also an eigenvector of $A$. So $A(\lambda x)=\lambda^2 x$. Since we assumed $\lambda$ is nonzero, $\lambda^2$ will be nonzero and thus $\lambda^2 x$ is nonzero. But this contradicts our assumption that $A(\lambda x)=0$ since $x$ is a nonzero eigenvector. So the only eigenvalue $A$ can have is zero.

QED

Is this a good rigorous proof of the statement the problem gives?

Thomas Andrews
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X Stanton
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1 Answers1

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As scalars commute with linear transformations, just note $$ 0=A^2x=A\lambda x=\lambda Ax=\lambda^2 x$$ and from $x\ne 0$ conclude $\lambda^2=0$ and finally $\lambda=0$

Hagen von Eitzen
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    Ok. But I'm a bit confused. My initial question was about if my proof was correct and rigorous. If this is as simple as it had to be, can a proof be too rigorous or redundant? Thanks. – X Stanton Jun 21 '21 at 19:37