# Nilpotent operator / Orthogonal projection

If you have a nilpotent operator $A \colon V \to V$, $V$ being $A$-cyclic (meaning that $V$ is generated by a single vector $v$ in $V$), is it true that the minimal polynomial of the Gram operator $P = A^{T} A$ is $p(t) = t^2 - t$, i.e., $P$ is the orthogonal projection?

-
Do you have any normalizing condition on $A$ (for instance norm one)? Because otherwise if one $A$ satisfies your condition, so will a scalar multiple and you wouldn't get a projection. – Martin Argerami May 29 '13 at 1:52
You're right, this isn't true under the given conditions. – Br09 May 29 '13 at 2:08

If you grant Jordan canonical form, then in a suitable basis $A$ looks like $$\begin{bmatrix}0&1&0&\dots&0 \\ 0&0&1&\dots&0 \\ & & \ddots&\ddots \\ 0&0&\dots&0&1 \\ 0&0&\dots &0&0 \end{bmatrix}\,.$$ Then $$A^\top A = \begin{bmatrix}0 & \\ & 1 & \\ & & \ddots \\ & & & 1\end{bmatrix}\,.$$ So 'twould appear you're right.
But .... Note that once you bring the transpose into the game, things are no longer basis-independent. The basis that brings $A$ into this nice form is not likely to be orthonormal, and so, if you're bringing the dot product into the game, it really is not meaningful to describe this in terms of an orthogonal projection. The characteristic polynomial of $A$ is basis-independent, but the characteristic polynomial of $A^\top A$ is not.