Part (a): By definition, the null space of the matrix $[L]$ is the space of all vectors that are sent to zero when multiplied by $[L]$. Equivalently, the null space is the set of all vectors that are sent to zero when the transformation $L$ is applied. $L$ transforms all vectors in its null space to the zero vector, no matter what transformation $L$ happens to be.
Note that in this case, our nullspace will be $V^\perp$, the orthogonal complement to $V$. Can you see why this is the case geometrically?
Part (b): In terms of transformations, the column space $L$ is the range or image of the transformation in question. In other words, the column space is the space of all possible outputs from the transformation. In our case, projecting onto $V$ will always produce a vector from $V$ and conversely, every vector in $V$ is the projection of some vector onto $V$. We conclude, then, that the column space of $[L]$ will be the entirety of the subspace $V$.
Now, what happens if we take a vector from $V$ and apply $L$ (our projection onto $V$)? Well, since the vector is in $V$, it's "already projected"; flattening it onto $V$ doesn't change it. So, for any $x$ in $V$ (which is our column space), we will find that $L(x) = x$.
Part (c): The rank is the dimension of the column space. In this case, our column space is $V$. What's it's dimension? Well, it's the span of two linearly independent vectors, so $V$ is 2-dimensional. So, the rank of $[L]$ is $2$.
We know that the nullity is $V^\perp$. Since $V$ has dimension $2$ in the $4$-dimensional $\Bbb R^4$, $V^\perp$ will have dimension $4 - 2 = 2$. So, the nullity of $[L]$ is $2$.
Alternatively, it was enough to know the rank: the rank-nullity theorem tells us that since the dimension of the overall (starting) space is $4$ and the rank is $2$, the nullity must be $4 - 2 = 2$.
The formula you mentioned is about projections on vectors. The problem here is about projections on spaces.
Determine an orthogonal basis $\{e_1,e_2\}$ of the space spanned by the collumns, using Gram-Schmidt. Then the projection of $b$ is $\langle b,e_1\rangle e_1+\langle b,e_2\rangle e_2$.
Best Answer
Let's first consider the orthogonal projection case and find its matrix $P$. We have that $$ A = \begin{bmatrix}1\\2\end{bmatrix} $$ is a basis for the range of the projection and then have $$ P = A(A^\top A)^{-1}A^\top = \frac15 \begin{bmatrix}1\\2\end{bmatrix} \begin{bmatrix}1 & 2\end{bmatrix} = \begin{bmatrix}1/5 & 2/5\\2/5 & 4/5\end{bmatrix}. $$ Now let's find the oblique projection $Q$ through $V=[7\;4]^\top$. Now let $$ B = \begin{bmatrix}-4\\7\end{bmatrix} $$ be a basis for the orthogonal complement of the null space of the projection We then have $$ Q = A (B^\top A)^{-1} B^\top = \frac1{10}\begin{bmatrix}1\\2\end{bmatrix} \begin{bmatrix}-4 & 7\end{bmatrix} = \frac{1}{10} \begin{bmatrix}-4 & 7\\-8 & 14\end{bmatrix}. $$ Then we can see that, for example, $Q$ projects the point $\begin{bmatrix}7\\-10\end{bmatrix}$ to $\begin{bmatrix}-9.8\\-19.6\end{bmatrix}$.
Visually, what's happening is that we're looking for the intersection between the line $y=2x$ and the line (in point-slope form) starting at $(7,-10)$ with slope $4/7$, i.e., $y-(-10) = \tfrac47(x-7)$. Seen another way, we're looking to solve the system
\begin{align} \begin{bmatrix} 7 \\ -10 \end{bmatrix} + c\begin{bmatrix} 7 \\ 4 \end{bmatrix} = d \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align} where $c$ and $d$ are constants that tell us "how far to go" for the lines to intersect.
We use the specific form of $B$ that is the orthogonal complement of $\begin{bmatrix} 7 \\ 4 \end{bmatrix}$ since this normal vector defines the hyperplane (or line) $\begin{bmatrix} 7 \\ 4 \end{bmatrix}$ whose direction we want to project along.