Find the standard matrix and kernel for a linear transformation. Let $T : \mathbb{R^3} → \mathbb{R^3}$ be a linear transformation given by $T(u) = \operatorname{proj}_vU$ where $v = (2, 0,−3)$.
(a) Find the standard matrix for $T$.
(b) Find a basis for the kernel of $T$.
I am completely lost on this particular question...I am familiar on finding the standard matrix and kernel but this question is a bit different...For instance, for finding Kernel of $T$, 2 vectors should be given, '$v$' and '$u$', but in this question, only '$v$' is given... 
 A: A more computationally intensive approach:
Note that the vectors $x_1=(3,0,2), x_2=(0,1,0)$ are orthogonal to $v$, hence 
$Tx_k = 0$. Also, $Tv = v$.
Hence in the ordered basis $v,x_1,x_2$, the operator $T$ has the form $\operatorname{diag}(1,0,0)$.
If we let $B=\begin{bmatrix} v & x_1 & x_2 \end{bmatrix}$, we have 
$T = B \operatorname{diag}(1,0,0) B^{-1} $.
Performing the computations gives
$T = {1 \over 13}\begin{bmatrix} 4 & 0 & -6 \\
0 & 0 & 0 \\ -6 & 0 & 9\end{bmatrix}$.
The kernel is easy to compute from the first line.
A: Hint
$T : \mathbb{R^3} \to \mathbb{R^3}$ can be written as 
$$T(x,y,z)=\frac{2x-3z}{13}(2,0,-3)=\left(\frac{4x-6z}{13},0,-\frac{6x-9z}{13}\right),$$ just making use of the definition of $\mathrm{proj}_{(2,0,-3)}(x,y,z).$ Since you say you are familiar with this, I think you can get the solution from this point.
If you get some geometric intuition of the problem it would be helpful. $T$ projects any vector on the $3$-dimensional space over a line through the origin. What happens if we project orthogonally a perpendicular vector to the line? And if the vector is parallel to the line? 
