Jtdylktuy

So..I have to find any linear map $f: mathbb{R}^4 rightarrow mathbb{R}^3$ that has a kernel and and an image with the following basis:

$$ker(f)=operatorname{Span}{(-1,0,0,1),(1,3,2,0)}.$$

$$operatorname{Img}(f)=operatorname{Span}{(1,1,1),(0,-2,1)}.$$

I have been trying to find similar problems to this one, but all of them so far have been for linear maps on $mathbb{R}^3 rightarrow mathbb{R}^3$, which are solved with Gauss. I have also seen that this can be solved by the extension of basis, but I can't wrap my mind around it as we have not touched that at all. What should be the method for solving this?

Lets fix: $V:= mathbb{R}^4, K:= leftlangle begin{pmatrix}-1 \ 0\0\1end{pmatrix} ,begin{pmatrix}1 \ 3\2\0end{pmatrix} rightrangle , I:= leftlangle begin{pmatrix}1 \ 1\1 end{pmatrix} ,begin{pmatrix} 0\-2\1 end{pmatrix} rightrangle , W:= mathbb{R}^3$

Now clearly: $K subset V$ and $I subset W$, this means we have canonical maps: $pi:Vtwoheadrightarrow V/_K$ and $iota:I hookrightarrow W$ (the projection onto the quotient and the inclusion). Now by the dimension formula we know $dim( V/_K)=2=dim(I)$, hence there exists an isomorphism $varphi:V/_K to I$ (pick your favourite one).

Consider the morphism: $$iota circ varphi circ pi: Vtwoheadrightarrow V/_K xrightarrow{sim}I hookrightarrow W.$$

Now since both, $varphi$ and $iota$ are monics, the kernel of $iota circ varphi circ pi$ is the same as the kernel of $pi$ which construction is $K$. Dually since $varphi$ and $iota$ are epics, the image of $iota circ varphi circ pi$ is the same as the image of $iota$ which by construction is $I$.

So $iota circ varphi circ pi$ has the desired properties

Now a funfact at the end: by the homomorphism theorem any morphism with the desired properties factors in precisely that way and "only" depends on the choice of $varphi$.

we need a $3times 4$ matrix.

The columns must be linear combinations of the vectors in the image.

let start with the first 2 columns, as the two given vectors.

$begin {bmatrix} 1&0&a&b\1&-2&c&d\1&1&e&f end{bmatrix}$

And our kernel maps to 0.

$begin {bmatrix} 1&0&a&b\1&-2&c&d\1&1&e&f end{bmatrix} begin {bmatrix} -1&1\0&3\0&2\1&0end{bmatrix} = begin {bmatrix} 0&0\0&0\0&0\0&0end{bmatrix}$

And that is enough to solve for the missing values.

$begin {bmatrix} 1&0&-frac 12&1\1&-2&frac 52&1\1&1&-2&1 end{bmatrix}$

We find a $3$ by $4$ matrix $$A= [C_1, C_2, C_3, C_4]$$ to represent our linear map.

From the information about the Kernel, we know that $$-C_1+C_4=0$$ and $$C_1+3C_2+2C_3 =0$$

And from the Image we know that the Column space needs to span $(1,1,1)^T$ and $(0,-2,1)^T$

Now we form a matrix $A$ with the above conditions.

$$ A = begin{bmatrix}-3&1&0&-3\1&1&-2&1\-5&1&1&-5end{bmatrix}$$