# Why can't we find a standard matrix representation in terms of eigenvalues with only ones on the diagonal?

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?

• 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 $c$. – Somos Jan 22 at 12:32