There are several basis you can choose for a vector space.
Say $M$ is your matrix. Then $M\,\mathbb R^4$ is a vector space and since $\det(M)\neq 0$ it has dimension 4, that is any of its bases has four vectors.
Notice that any $v$ in this space can be written as $v=Mw$ for some $w$ so $v$ is a linear combination of the columns of the matrix, which means that the four vector columns of $M$ are a basis for the space.
Now say $\hat{M}$ is $M$ is in reduced column echelon form. Then the four vector columns of $\hat{M}$ are also a basis for the space.
We also have that $\mathbb{R}^4\, M$ is a vector space of dimension 4 since again the determinant of $M$ is not zero. Look at the space as the set of vectors $v$ such that $v=w \,M$ for some $w\in\mathbb{R}^4$. Any such $v$ is a linear combination of the row vectors of $M$. The same comments I made above with the respect the column echelon form also applies here with the row echelon form. The four row vectors of the row echelon matrix will be a basis for $\mathbb{R}^4\, M$.
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 $4 - 2 = 2$.
Best Answer
From here.
Wiki seems to say the same. It also says here that
I'm guessing that