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So I have proved that $0$ is an eigenvalue for the above matrix and that the geometric multiplicity of $0$ is $n-1$. I know if I can find one more eigenvector for some other eigenvalue, I will be able to prove this. But I don't know how to find this other eigenvector.

Novice
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Gureet
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2 Answers2

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If $\operatorname{rank}(A)=1$ then $A=vw^T$ for some non-zero vectors $v,w$. Then $Av=vw^Tv=\langle w,v\rangle v$.

Hagen von Eitzen
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Hint: If $x$ is in the column space of $A$, then $Ax$ must also be in the column space of $A$.

Ben Grossmann
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  • I understand this. But how does this makes finding the other vector any easy? How can I use this to get an answer? – Gureet Sep 23 '20 at 12:44
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    What do you mean by "the other vector"? In any case, my point is that the rank of $A$ is $1$. So, the column space is one dimensional. So, if $x$ is a non-zero element of the columns space and $Ax$ is in the column space, then $Ax$ is a multiple of $x$. – Ben Grossmann Sep 23 '20 at 13:04