(i) Suppose $A\in M_{9\times 4}(\mathbb{F})$ has a 4 dimensional kernel. Find $\mathrm{rank}(A)$.
My solution: By the rank-nullity formula we have $\dim(\ker A)+\mathrm{rank}(A)=\dim(A)$. So $\dim(A)=9, \dim(\ker A)=4$. Hence rearranging we have $\mathrm{rank}(A)=5$
(ii) Consider $\begin{pmatrix}1&1&1&3\\1&1&2&4\\1&1&1&3\end{pmatrix}$ to be a linear map from $\mathbb{F}^4$ to $\mathbb{F}^3$. Describe the kernel and image by finding a basis for each.
I struggled with this a lot more than the first problem. I started by saying that the kernel is the set $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$ satisfying $A\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\bf{0}$. We obtain 2 simultaneous equations:
$$a+b+c+3d=0$$
$$a+b+2c+4d=0$$
Subtracting gives $a+b=2c$ and $c=-d$. So a general solution is $(a,2c-a,c,-c)$. This doesn't get me any closer to a solution though.
Can someone comment on these solutions and help with my attempt for (ii) in particular?
Best Answer
By the definition of rank (the maximun number of linearly independent columns or rows), you can figure out right away $\operatorname{rank} (A)$, and by the rank-nullity theorem you can calculate $\operatorname{null}(A)$. To find a basis of each then should be easy then, you are very close to getting one for $\ker(A)$: if the general solution to that system is $(a,2c−a,c,−c)$, you can write it as $a \cdot (1, -1, 0, 0)+ c\cdot (0, 2, 1, -1)$, and since $a$ and $c$ are arbitrary, that is nothing more than a linear combination of the two vectors $(1, -1, 0, 0),(0, 2, 1, -1)$ which then form a basis. To figure out a basis for the image, remember that the columns of the matrix taken as vectors span the image itself.