find adjoint of a linear transformation defined by a cross product Suppose that a in R^3 is given. Define A in L(R^3) by Av = a (cross)v
find adjoint A, assuming that R^3 is equipped with the standard dot product
I know to find adjoint A, I must start from definition (Av, y) = (v, A* y) 
However, when replace Av = a x v, I don't know how to do more
((a x v), y)= (a,y)x (v,y) 
Someone helps me, please
 A: You can solve the problem without computations by geometrical reasoning. The triple product 
$$
\langle a\times v , y\rangle
$$
equals the signed volume of the parallelogram spanned by the vectors $a, v, y$, in that order. To write it in adjoint form you need to fill in the blank in the following formula: 
$$
\langle a\times v , y\rangle=\langle v, ?\rangle .$$
If you set $?=y\times a$ then you get for the left hand side the triple product of $v, a, y$, that is, the signed volume of the same parallelogram as before BUT with two sides permuted. Therefore, the sign changes and the correct answer is 
$$
\langle a\times v , y\rangle=-\langle v, a\times y\rangle .$$
(Of course you could have arrived at this result with a more algebraic approach, and probably if this is homework you'd better do so.)
A: Given a vector $a\in\mathbb R$, the matrix representation of the linear transformation $A:v\mapsto a\times v$ with respect to the standard basis, usually denoted by $[a]_\times$, is a skew-symmetric matrix known as the cross product matrix. Therefore the matrix of $A^\ast$ is $[a]_\times^T=-[a]_\times$, i.e. $A^\ast v=-av$.
A: computing Av = a x v
let a =(a1,a2, a3) and v = (v1,v2,v3) apply cross product, I got Av = a2v3-a3v2-a1v3+a3v1+a1v2-a2v1
under the matrix form, it is Av = {{0, -a2, a3},{a1,0,-a3},{{-a1,a2,0}}{v1,v2,v3}
so that A = [col1 col2  col3] as shown
A = A^t
