Let $a=(a1,a2,a3)$ be a fixed vector in $\Bbb R^3$. Define the cross product $a \times v$ of $a$ and another vector $v=(v_1,v_2,v_3) \in \Bbb R^3$ as $$a \times v = \det\begin{bmatrix} e_1 & e_2 & e_3 \\ a_1 & a_2 & a_3 \\ v_1 & v_2 & v_3 \\ \end{bmatrix}$$
Define a function $T: \Bbb R^3 \to \Bbb R^3$ by $T(v) = a \times v$ for $v \in \Bbb R^3$.
a) Show that $T$ is a matrix transformation and calculate its representing matrix $M$
b) Find $\operatorname{ker}(T)$ and interpret its answer geometrically
c) find $\operatorname{range}(T)$ and interpret its answer geometrically
im really lost on this question. how do you find matrix M? i guess in order to find KerT you have to find matrix M first, and then find null space of it. and rangeT.. i dont know about that either..
$T:\Bbb R^3 \to \Bbb R^3$ is given by $T(\mathbf v) = \mathbf a \times \mathbf v = \det\begin{bmatrix} \mathbf e_1 & \mathbf e_2 & \mathbf e_3 \\ a_1 & a_2 & a_3 \\ v_1 & v_2 & v_3 \\ \end{bmatrix}$ where $\{\mathbf e_1, \mathbf e_2, \mathbf e_3\}$ are the standard basis vectors for $\Bbb R^3$. For instance, $\mathbf e_3 = (0,0,1)$.
Because $T$ is a linear transformation, if we can figure out how it transforms any basis then we can figure out how it transforms any vector -- and more importantly for this question, what its standard matrix is.
So let's see how $T$ transforms the standard basis (being the easiest one to check, usually).
$$\begin{align}T(\mathbf e_1) = \mathbf a \times \mathbf e_1 &= \det\begin{bmatrix} \mathbf e_1 & \mathbf e_2 & \mathbf e_3 \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{bmatrix} \\ &= \det\begin{bmatrix}a_2 & a_3 \\ 0 & 0\end{bmatrix}\mathbf e_1 - \det\begin{bmatrix}a_1 & a_3 \\ 1 & 0\end{bmatrix}\mathbf e_2 + \det\begin{bmatrix}a_1 & a_2 \\ 1 & 0\end{bmatrix}\mathbf e_3 \\ &= a_3\mathbf e_2 -a_2\mathbf e_3\end{align}$$
Likewise, you can confirm for yourself that $$T(\mathbf e_2) = -a_3\mathbf e_1+a_1\mathbf e_3 \\ T(\mathbf e_3) = a_2\mathbf e_1 -a_1\mathbf e_2$$
Then we just need to find the matrix $M$ such that $M\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix} 0 \\ a_3 \\ -a_2\end{bmatrix}$ and the similar expressions for the other basis vectors.
But $M\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}$ is just the first column of $M$ (Can you see why?). Thus $$M=\begin{bmatrix} M\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} & M\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} & M\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\end{bmatrix} = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0\end{bmatrix}$$
Here's how I'd find the kernel and range of $T$.
From the properties of the determinant we can see that $\mathbf a\times \mathbf v=\mathbf 0$ iff $\mathbf a \| \mathbf v$. Thus $\ker(T) = \Bbb R^3$ if $\mathbf a=\mathbf 0$ or $\ker(T) = \operatorname{span}(\mathbf a)$ if $\mathbf a\ne \mathbf 0$.
Then we can use the rank-nullity theorem to get an idea of the range of $T$. If $\mathbf a=\mathbf 0$ then we can immediately see that $\operatorname{range}(T) = \{\mathbf 0\}$. If $\mathbf a\ne \mathbf 0$, then the range is $2$-dimensional.
Let's assume $\mathbf a\ne \mathbf 0$ and consider the dot product of $\mathbf a\times \mathbf v$ with $\mathbf a$.
$$\mathbf a \cdot (\mathbf a\times \mathbf v) = (a_1\mathbf e_1 + a_2\mathbf e_2 + a_3 \mathbf e_3)\cdot \left(\det\begin{bmatrix}a_2 & a_3 \\ v_2 & v_3\end{bmatrix}\mathbf e_1 - \det\begin{bmatrix}a_1 & a_3 \\ v_1 & v_3\end{bmatrix}\mathbf e_2 + \det\begin{bmatrix}a_1 & a_2 \\ v_1 & v_2\end{bmatrix}\mathbf e_3\right) \\ = \det\begin{bmatrix} a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3 \\ v_1 & v_2 & v_3\end{bmatrix} = 0$$
Thus $T(\mathbf v)\ \bot\ \mathbf a$. But we know that $\operatorname{span}(\mathbf a)^\bot$ is $2$-dimensional, thus this must be the range of $T$.
Geometrically speaking this means that $\ker(T)$ is the line parallel to $\mathbf a$ and $\operatorname{range}(T)$ is the plane orthogonal to that line.