If $A$ is your matrix, the null-space is simply put, the set of all vectors $v$ such that $A \cdot v = 0$. It's good to think of the matrix as a linear transformation; if you let $h(v) = A \cdot v$, then the null-space is again the set of all vectors that are sent to the zero vector by $h$. Think of this as the set of vectors that lose their identity as $h$ is applied to them.
Note that the null-space is equivalently the set of solutions to the homogeneous equation $A \cdot v = 0$.
Nullity is the complement to the rank of a matrix. They are both really important; here is a similar question on the rank of a matrix, you can find some nice answers why there.
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
Dear fellow mathematician,
at first, it might appear that it is not really as useful. But the opposite is true, you are going to use it to find eigenvectors which are of huge importance in linear algebra. They are used for diagonalising matrices and for Singular Value Decomposition which is of vital importance. (it’s a numerically stable way of taking powers of matrices - solving differential equations and many other things - i.e. finding n-th term of a sequence from a reccurent formula, which can be used for various)
These are the applications that I’m familiar with, I’m sure that someone will provide more.
Have a nice Sunday!