# Transformation matrix is jordan normal form

I have the following question: Given a finite-dimensional, unitary vector space V and a endomorphism f on V, is it possible to choose an orthonormal basis B of V in such a way, that the transformation Matrix $M_{B}^{B}(f)$ is the jordan normal form of f?

I would say yes, because f is triangularisable on V and therefore the jordan normal form of f exists. Unfortunately I have no idea, how to start this proof. Thus I would appreciate it, if you could give me a hint. Thanks in advance!

Your statement is false, with or without normalization. For a "trivial" counter-example, take $V$ 2-dimensional with some orthonormal basis $\{e_1, e_2\}$, and define $f e_1 = 7 e_2$, $f e_2 = 0$ (so $f^2 = 0$). In this case (and for all 2D vector spaces) there is an orthogonal, although not necessarily orthonormal basis where the matrix of $f$ has its Jordan normal form.
For a more non-trivial counter-example where orthogonality is also violated, take $V$ 3 dimensional, with some orthonormal basis $\{e_1, e_2, e_3\}$, and define $f$ by $f e_1 = e_1 + e_2$, $f e_2 = -e_1 - e_2 + e_3$, and $f e_3 = 0$. It is easy to check that the matrix of $f$ is in its Jordan normal form in the non-orthonormal $\{e_1, e_1 + e_2, e_3\}$ basis, and further that $f^2 e_1 = - f^2 e_2 = e_3$. Thus an element proportional to $e_3$ must be present in any basis where the matrix of $f$ has its Jordan normal form. Suppose then that $f$ has Jordan normal form with respect to some basis $\{v_1, v_2, \alpha e_3\}$ for some unknown constant $\alpha \ne 0$ and $\langle v_1, e_3 \rangle = 0$. Relabeling if necessary, we then have $f v_1 = v_2$, but if $v_1$ has any component along $e_2$, then $\langle v_2, e_3 \rangle \ne 0$, so $v_1 \propto e_1$. But then $\langle v_1, v_2 \rangle \ne 0$.