1. ## Linear maps

Consider the linear map defined on $R^4$ with values in $R^3$ given by the matrix:
A =
( 1 3 1 -2 )
( 3 9 4 -4 ) yes, this is supposed to be a matrix, i'm new
(-1 -3 -2 0 )
a) find all x e $R^4$ satisfying the linear equation Ax=b, with
b =
( 1)
(-2)
( 4)

b) find a basis for ker A

c) Using the Rank Theorem, find the dimension of im A

This is a past paper without solutions and would really help me revise, thank you very much !

2. Well, I don't know if there is an easier way than this, but I suspect
there is:

a) If you multiply Ax=b then you will get three equations for which
3 of the variables in x can be eliminated leaving you with

\begin{aligned}
x & =6+2t\\
y & =\frac{2}{3}t\\
z & =-5-2t\\
t & =t\end{aligned}

so \begin{aligned}
\mathbf{x} & =\left(\begin{array}{c}
6+2t\\
\frac{2}{3}t\\
-5-2t\\
t\end{array}\right)\end{aligned}

b) the kernel of A can be found in a similar way Ax=0

\begin{aligned}
\mathbf{ker(A)} & =\left(\begin{array}{r}
0\\
\frac{4}{3}t\\
-2t\\
t\end{array}\right)=\frac{t}{3}\left(\begin{array} {r}
0\\
4\\
-6\\
3\end{array}\right)\end{aligned}

which is a straight line through the origin,

so a basis could be $\left(\begin{array}{r}
0\\
4\\
-6\\
3\end{array}\right)$
which has dimension 1, ie n(A)=1

c) I'm not sure what the rank theorem is but r(A)+n(A)=dimX means
that r(A)+1=4 so n(A)=3. But I thought that was kind of obvious anyway,
so maybe I misunderstood the question.

3. The "rank theorem" does, indeed, say that, if A is a linear operator from vector space U to vector space V, then the rank of A plus the nullity (dimension of the kernel) of A is equal to the dimension of U. And you have a typo:
r(A)+ n(A)= 4 and n(A)= 1 so r(A)= 3, not "n(A)= 3". That says that A maps $R^4$ onto $R^3$ which was not entirely "obvious". The rank of A, the dimension of its image in $R^3$ must be less than or equal to 3 but certainly might have been less.

4. Ah, I see, thanks for clearing that one up HallsofIvy.