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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?

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  • $\begingroup$ what do you mean with find? I am currently tempted to solve it with universal properties (the solution can't be unique, hence I could at least try some canonicality). i.e. if you want a specific solution, you would have to trace that all back $\endgroup$ – Enkidu Jan 8 at 10:38
  • $\begingroup$ Sorry, I edited it now, I meant find as in finding any linear map that meets those criteria, so a universal solution would work too, as would do any specific solution. $\endgroup$ – Lightsong Jan 8 at 10:51
  • $\begingroup$ ok, then I will write something down: however, be aware that i won't be a universal solution, but its construction will involve universal properties, which makes it in a way "canonical" $\endgroup$ – Enkidu Jan 8 at 10:52
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Lets fix: $V:= \mathbb{R}^4, K:= \left\langle \begin{pmatrix}-1 \\ 0\\0\\1\end{pmatrix} ,\begin{pmatrix}1 \\ 3\\2\\0\end{pmatrix} \right\rangle , I:= \left\langle \begin{pmatrix}1 \\ 1\\1 \end{pmatrix} ,\begin{pmatrix} 0\\-2\\1 \end{pmatrix} \right\rangle , W:= \mathbb{R}^3$

Now clearly: $K \subset V$ and $I \subset W$, this means we have canonical maps: $\pi:V\twoheadrightarrow V/_K$ 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: $$\iota \circ \varphi \circ \pi: V\twoheadrightarrow V/_K \xrightarrow{\sim}I \hookrightarrow W.$$

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 $\varphi$.

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    $\begingroup$ In other words, you showed that there exists a canonical bijective correspondence between the set $S(K,I)\subseteq \operatorname{Hom}(V,W)$ of all linear maps $T:V\to W$ such that $\ker T=K\subseteq V$ and $\operatorname{im} T=I\subseteq W$ and the set $\operatorname{Iso}(V/K,I)$ of isomorphisms from $V/K$ to $I$. In the case $V/K\cong I$, the set $\operatorname{Iso}(V/K,I)$ is (non-canonically) in bijective correspondence with $\operatorname{GL}(V/K)$, or with $\operatorname{GL}(I)$. Otherwise, $\operatorname{Iso}(V/K,I)$ is empty. Great work! $\endgroup$ – user593746 Jan 8 at 12:45
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    $\begingroup$ actually not, that is the homomorphism theorem, I just did the other direction: sazing that a morphism that factors that way has the desired properties and stating that they exist by dimension. $\endgroup$ – Enkidu Jan 8 at 13:03
  • $\begingroup$ Wow...That is quite out of my league, but it's an amazing answer, I will be checking this one again once I finish this year with,hopefully, more knowledge of linear Algebra. Thank you for the answer and sorry for the delay in the response. $\endgroup$ – Lightsong Jan 11 at 11:29
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we need a $3\times 4$ 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.

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

And that is enough to solve for the missing values.

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

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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.

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

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