# Cayley Hamilton Theorem proof

The last step in todays lectures proof of the cayley hamilton theorem doesn't make sense to me. No problems up until the point where we showed that the characteristic polynomial we are trying to prove is the zero function is equal to: q(T)*p(T) where p(T) is the zero function.

Our professor then finishes the proof with the following steps: q(T)*p(T)(v)= q(T) performed on p(T)(v) which is equal to zero for any v since p(T)(v) is zero and q(T) beeing a linear function maps zero to zero. What I dont understand is the step q(T)*p(T)(v)= q(T) performed on p(T)(v). Wouldnt it be correct to finsh the proof saying q(T)*p(T)(v)= q(T)(v)*p(T)(v)=q(T)(v)*0=0?

• The problem is that $\underbrace{ (q(T)p(T))(v)}_{vector}=\underbrace{ q(T)(v)}_{vector}*\underbrace{ p(T)(v)}_{vector}$ doesn't make sense: what would be the meaning of the "*" operation : dot product ? but in this case, the result would be a scalar... – Jean Marie Apr 21 '17 at 22:31

In your last sentence, $q(T)* p(T)(v) = q(T)(v) * p(T)(v)$ does not actually make sense. $q(T)$ and $p(T)$ are linear operators, so $q(T)(v)$ and $p(T)(v)$ are vectors, for which multiplication does not make sense.
Let's just call $q(T)$ and $p(T)$ by $Q$ and $P$.
What your professor is saying is that $QP(v) = Q(Pv)$. In general, this is the defining property of an algebraic action on a set. It doesn't matter if you multiply $Q$ and $P$ in operator space and then act on $v$, or if you act on $v$ by $P$ and then act on the result of that by $Q$: you get the same thing either way!
Then, as you said, $Pv$ is $0$ since, $P$ is the $0$ operator, and Thus $Q(Pv) = 0$ since linear operators map $0$ to $0$.