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(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?

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  • $\begingroup$ I've never seen someone write $dim(A)$. It should be the dimension of the domain of $A$, and since $A\in M_{9\times4} (\mathbb F) $, the domain is... For the second problem, you can proceed as you did to figure out the dimension of the kernel and one of its bases, and to find the rank you should simply use its definition. Have you studied gaussian elimination? $\endgroup$ Sep 1 '21 at 14:25
  • $\begingroup$ @blundered_bishop Since $A\in M_{9\times 4}(F)$ surely the domain is $9$? I've studied gaussian elimination but not sure how it applies here since we have only 2 distinct equations but 3 matrix rows $\endgroup$
    – user933413
    Sep 1 '21 at 14:27
  • $\begingroup$ The domain is indeed $9$, sorry if my wording suggests otherwise. Gaussian elimination can be used to solve any linear system, even when some rows are similar: you are going to end up with a row of only zeroes. I'll post the rest as an answer because it is too long for a comment. $\endgroup$ Sep 1 '21 at 14:50
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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.

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  • $\begingroup$ Thanks! This is a really useful answer:) $\endgroup$
    – user933413
    Sep 1 '21 at 15:01

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