So I wanted to calculate something like $\langle Ax,y\rangle$. Is there some way to take the A out so I have sth like $f(A)\langle x,y\rangle$? Where $f(A)$ is some kind of transformation of A.
Scalar product with a matrix coefficient
1
$\begingroup$
matrices
vectors
1 Answers
5
In general, no.
For instance, if $A$ represents rotation in the euclidean plane through 90 degrees, and $x$ and $y$ are perpendicular unit vectors, then $\langle Ax, y \rangle$ will always be $\pm 1$, but $\langle x, y \rangle = 0$. There's nothing you can multiply $0$ by to get $\pm 1$.
Let me go further: not only is it not true for the example I gave, it's essentially almost never true.
Suppose that for a matrix $A$, there's some constant $c$ such that $\langle Ax, y \rangle = c \langle x \rangle y$.
For any nonzero vector $v$, let $P(v)$ be the set of all vectors perpendicular to $v$. Consider a vector $w \in P(v)$. We have
$\langle Aw, v \rangle = c \langle w \rangle v$ = 0.
So $A$ takes elements in $P(v)$ into $P(v)$.
In particular, for each standard basis vector $e_i$, let $P_i$ denote the hyperplane consisting of vectors whose $i$th entry is zero, i.e., $P(e_i)$. Then we've shown that $AP_i \subset P_i$. That means that $$ A(P_i \cap P_j) \subset P_i \cap P_j, $$ and similarly for all higher order intersections. In particular, it means that $A$ must map each standard vector to a multiple of itself. And that multiply must (by the assumed property of $A$) be he number $c$. Hence $$ A = cI. $$
In short: the only matrices with the property you seek are scalar multiples of the identity.
(There's probably some one-liner proof of this using Schur's Lemma or something...but this proof is self-contained and elementary, which appeals to me.)