For every linear map $M: V \to W$ between real finite dimensional vector spaces one can always choose a basis of $W$ and one of $V$ so that the matrix representation of $M$ has $n :=$rank$(M)$ ones along the diagonal and only zeros elsewhere. Now let $V = W$, we can compute its eigenvalues and the standard form of its matrix representation would contain only non-zero eigenvalues of the diagonal. Why can we not change the basis to make all the entries equal to one, as before?

I thought that if we choose a basis of eigenvectors, we would have have the eigenvalues of the diagonal (as i.e. in Jordan normal form). Now, if we scale those eigenvectors with their eigenvalues, we should have ones along the diagonal, but why isn't that possible?

  • $\begingroup$ Suppose $V$ is one dimensional. What does your question mean in this special case? Remember any linear map on $V$ is $x \mapsto cx$ for some scalar $c$. $\endgroup$ – Somos Jan 22 at 12:32

Because now we only have one basis to deal with, whereas when we are dealing with two vector spaces we can deal with two bases, one for each space.

  • $\begingroup$ Why can't we choose a two different basis for the "domain"- and "target"-vector space? $\endgroup$ – Viktor Glombik Jan 22 at 12:14
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    $\begingroup$ We can do that. Usually, it is not done and the statement (in bold) that you are trying to prove employs the singular “basis” and not the plural “bases”. $\endgroup$ – José Carlos Santos Jan 22 at 12:18

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