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:



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?

  • $\begingroup$ 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

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 $\varphi$.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

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


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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.