$A$, $B$ have all $\lambda\geq 0$, null($A$)=null($A^2$), null($B$)=null($B^2$), $A^4=B^4$, then $A=B$.

Asked Oct 23 '23 at 17:48

Active Oct 23 '23 at 17:54

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Problem

Let $A,B\in M_n(\mathbb C)$. Suppose $\text{null}A=\text{null}A^2,\text{null}B=\text{null}B^2$ and the spectrums of $A$ and $B$ are nonnegative. If $A^4=B^4$, then $A=B$.

Similar Question

$A$, $B$ have all $\lambda \geq 0$, null($A$) $=$ null($A^2$), null($B$) $=$ null($B^2$). Prove that if $A^2 = B^2$, then $A=B$.

I know that if I can solve the similar question above, then this problem is just a straightforward corollary. However, the answer in the link, as far as I'm concerned, is quite vague; so I am here craving for an extra explanation.

edited Oct 23 '23 at 17:54

user26857

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asked Oct 23 '23 at 17:48

SuperSupao

1

You should edit your post to specify what $\lambda$ is. – Digitallis Oct 23 '23 at 19:15
I've marked that as a duplicate of a question asked last month. You should know that original posters are expected to show some effort. – user8675309 Oct 23 '23 at 19:29

$A$, $B$ have all $\lambda\geq 0$, null($A$)=null($A^2$), null($B$)=null($B^2$), $A^4=B^4$, then $A=B$.

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