Linear Transformation - Bonus Question Set Let $\underline v = (v_1, v_2, v_3)∈\mathbb{R}^3$ and $\underline w = (w_1, w_2)∈\mathbb{R}^2$ be non-zero row vectors. 
Define $T(\underline x) = (\underline v \cdot\underline x)\underline w$
(a) Show that $Kernel$$(T)$ $= (span${$\underline v$}$)^\bot$
(b) Show that $Range(T)=span${$\underline w$}
(c) Show that the standard matrix of $T$ is $\underline w ^t \underline v$
This is a bonus question set we got in class late last week. I'm going to run over the solution with my tutor on Tuesday, however I'd love to get my head around it prior to that. 
I know that for the last part I would need to show that $T(\underline x)=$$(\underline w ^t \underline v) \underline x$, as $T(\underline x)=A\underline x$ where A is the standard matrix of the linear transformation T. 
Besides that, I have no idea how to go about solving this and any help would be most appreciated.
 A: You can start representing explicitly your transformation. Here I represent vectors as columns, so :
$$
\vec v= \begin {bmatrix}
v_1\\v_2\\v_3
\end{bmatrix}
\qquad and \qquad 
\vec x= \begin {bmatrix}
x\\y\\z
\end{bmatrix}
$$
And, by definition:
$$
T(\vec x)=(v_1x+v_2y+v_3z)
\begin {bmatrix}
w_1\\w_2
\end{bmatrix}
=
\begin {bmatrix}
(v_1x+v_2y+v_3z)w_1\\(v_1x+v_2y+v_3z)w_2
\end{bmatrix}=
$$
$$
=
\begin {bmatrix}
v_1 w_1x+v_2w_1y+v_3 w_1z\\v_1 w_2x+v_2 w_2y+v_3 w_2z
\end{bmatrix}=
\begin {bmatrix}
v_1w_1&v_2w_1&v_3 w_1\\v_1 w_2&v_2 w_2&v_3 w_2
\end{bmatrix}
\begin {bmatrix}
x\\y\\z
\end{bmatrix}=
$$
$$
\begin {bmatrix}
w_1\\w_2
\end{bmatrix}
\begin {bmatrix}
v_1&v_2&v_3
\end{bmatrix}
\begin {bmatrix}
x\\y\\z
\end{bmatrix}
$$
Since you have a notation that change columns-rows, this prove your claim (c) (you can see this?).
Now an inspection to the passages gives also the answers to the questions (a) and (b): from the first identity you see that $T(x)$ span the linear space of $\vec w$.
And you also see that $T(x)=0$ if the dot product $\langle \vec v, \vec x \rangle$ is null, i.e. if $\vec x$ is orthogonal to $\vec v$.
A: You should also try to get the concepts behind the calculations.
For question a), ask yourself, which vectors $x$ will be such that $T(x)=0$? Whatever vectors $x$ they may be, these are by definition ${\bf kern}(T)$. Now, we are looking for $x$ independent of $v$ and $w$. Clearly, the only option is that we take $x$ such that the dot product $(v\cdot x)=0$. In other words, we seek $x$ perpendicular to the vector space spanned by $v$ -it's just a 1-dimensional vector space. This is exactly what your are asked to prove: ${\bf kern}(T)=<span \,v>^\perp$. 
For question b),let's see, $T(x)$ gives just vector $w$ multiplied by a scalar $(v\cdot x)$. Thus, by varying $x$ we will run through all vectors parallel to $w$, i.e., ${\bf Range}(T)=<span\,w>$.
As for question c), let's first revisit what a linear function is. In terms of matrices, each column provides the image in the codomain of a base vector in the domain of the linear function. 
When you write $w^Tv$, a column vector on the left and a row one on the right, it means the direct product of the two vectors. The result is not a scalar but a matrix, where column $i$ consist of all the components of the column vector $w^T$ multiplied by the component $v^i$. (I'm assuming $T$ acts on the left, i.e., the action of $T$ on $x$ is $Tx\equiv T(x)$).
On the other hand, the standard matrix is given the image of the base vectors in the domain. Let's consider the case of $e_1=(1,0,0)$, its image is
the column given by $T(e_1)=v^1\,w$.This should more properly be written as $T(e_1)=v^1\,w^T$, as $w$ is a row vector by definition. 
Analogously, the image of the second base bector $e_2=(0,1,0)$ is $T(e_2)=v^2\,w^T$, and for the third base vector $e_3=(0,0,1)$ it is $T(e_3)=v^3\,w^T$. And these are the three columns of the matrix corresponding to $T$. That is, $T=w^Tv$.
