Nullity and Rank without a clear matrix For a fixed non-zero vector b =$\left(  
\begin{array}{c}
    b_1 \\
    b_2 \\
    b_3\\
  \end{array}
\right)$
, the mapping $\;T : \mathbb R^{3}\ \to\ \mathbb R^{3}\\;$
is defined by 
$T$(x) = x $\times$ b,
where $\times$ denotes the cross product.
i) Show that T is linear.
ii) Find nullity(T). Give a reason.
iii) Find rank(T). Give a reason.
iv) If b =$\left(  
\begin{array}{c}
    1 \\
    0 \\
    3\\
  \end{array}
\right)$
, find the matrix $A$ such that T(x) = $A$x.
So far, i) is shown with a zero vector and iv) matrix A is found, with matrix $A$ =$\left(  
\begin{array}{c}
    3 & 0 & 0\\
    -3 &0 & 1\\
    0 & -1 & 0\\
  \end{array}
\right)$
Well, not sure if iv) is right but thats what I got.
The problem is about ii) and iii), if either of them is found, then the other one will be found as well, I have got no idea how to find the nullity without a clear matrix...

Thank you.
 A: The nullity is the dimension of the kernel of $T$. Because $x \times b = 0$ whenever $x = \lambda b$ for an $\lambda \in \mathbb R$ the kernel of $T$ is the set $ker T = \{ \lambda b: \lambda \in \mathbb R\}$. This vector space is one dimensional and thus $\operatorname{nullity}(T)=1$.
A: Linearity results from the properties of the cross-product:


*

*$(u+u')\times v=u\times v+u'\times v$

*$(\lambda u)\times v=\lambda(u\times v)$

*$u\times (v+v')=u\times v+u\times v'$

*$u\times (\lambda v)=\lambda(u\times v)$

*$v\times u=-u\times v$
which can be summarised as  being antisymmetric bilinear.


The image of $T $ is the plane orthogonal to $\mathbf b$. It certainly is contained in that plane. Conversely, let $\mathbf u$ be a vector that is orthogonal to $\mathbf b$. Then, by the double cross product formula:
$$\mathbf b\times(\mathbf b\times \mathbf u)=\mathbf b(\mathbf b\cdot \mathbf u)-\mathbf u(\mathbf b\cdot \mathbf b)=-\mathbf u(\mathbf b\cdot \mathbf b),$$ 
so that $$\mathbf u=T\Bigl(\frac{\mathbf b\times \mathbf u}{\mathbf b\cdot\mathbf b}\Bigr).$$
Hence $\operatorname{rank}(T)=2$.

The matrix $A$ must be antisymmetric. The coordinates of $\mathbf x\times \mathbf b$ are given by the minors of the matrix:$$\begin{bmatrix}x_1&b_1\\x_2&b_2\\x_3&b_3 \end{bmatrix}$$
We find $x\mathbf=\begin{bmatrix}b_3x_2-b_2x_3\\b_1x_3-b_3x_1\\b_2x_1-b_1x_2\end{bmatrix} $, whence the matrix of $T$:
$$\begin{bmatrix}0&b_3&-b_2\\-b_3&0&b_1\\b_2&-b_1&0\end{bmatrix}=\begin{bmatrix}0&3&0\\ -3&0&1\\ 0&-1&0\end{bmatrix}$$
