I am given the following information:
Let v = (r, s, t) be a fixed vector in $\mathbb R^3$, and let C: $\mathbb R^3$-> $\mathbb R^3$, C(x) = v × x
I already proved that C is linear. Now, using the fact that C is linear and properties of the cross product, I am asked to show that
- T(x) = x × v is also linear
- Let v = (1, 2, -1). From properties of cross product infer the null space, column space, and rank of [C]
My attempt:
I know that a cross product in reverse order is just the negative of the original cross product, i.e, the cross product is anti-commutative. Is this information enough to conclude that T is linear? i.e., is it that simple?
As for the second part, I obtained the matrix for [C], but from there I'm lost. I am able to obtain the null space, column space, and rank of [C] by using row reduction of [C], but unfortunately they need it to be done using the properties of cross product.
Thank you for any help
You are right; the negative of a linear transformation is negative. Indeed, since $T(x)=-C(x)$ and $C(x)$ is linear we have $$T(x+y)=-C(x+y)=-[C(x)+C(y)]=-C(x)-C(y)=T(x)+T(y)$$ and $$T(cx)=-C(cx)=-cC(x)=cT(x)$$
The null space is the set of vectors whose cross product with $v$ is $\vec{0}$. You may have learned that the cross product of two vectors is $\vec{0}$ iff they are parallel. Therefore the null space is the set of all vectors parallel to $v$, which you can think of as all the vectors $c\langle 1,2,-1\rangle$ for $c$ any real number. The column space, or image, is the set of all vectors that can be expressed in the form $v\times x$ for some $x$. For this I would use the fact that the cross product of two vectors is perpendicular to both of them. Therefore the vectors that can't be expressed in that form are ones which are not perpendicular to $v$. Can you conclude?
Note that by Rank-Nullity Theorem the rank of $[C]$ is $2$ since we have shown the null space to have dimension $1$.