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Given that I have eigenvalues $\lambda_{1},\dots,\lambda_{n}$ and eigenvectors $v_{1},\dots,v_{n}$ for matrix $A$.

What would happen if I calculated the eigenvalues and eigenvectors of the squared matrix $A^2$?

Are they the squares of the previous values? Or are they totally unrelated?

I couldn't find this property online.

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Do you make the assumption that, in your case, there are as many distinct eigenvalues as the size of your matrix ? – LeGrandDODOM Jan 15 '14 at 16:41
up vote 1 down vote accepted


$$Av_i=\lambda_i v_i\Rightarrow A^2v_i=A(Av_i)=\lambda_iAv_i=\lambda_i^2v_i$$ so what you can conclude?

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So, the eigenvalues are squared and the eigenvectors are the same? – KuramaYoko Jan 15 '14 at 16:19
@KuramaYoko yes correct and you can generalize this for $A^p$. – user63181 Jan 15 '14 at 16:21

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