# Find a matrix of (any) linear map $\varphi : \mathbb{R}^{4} \rightarrow \mathbb{R}^{3}$, so that conditions apply

I'm trying to find a matrix of (any) linear map $$\varphi : \mathbb{R}^{4} \rightarrow \mathbb{R}^{3}$$, for which the following conditions apply:

• the dimension of the image $$\varphi = 2$$

• $$\varphi(1,1,1,1) = (1,2,1)$$

• $$\varphi(1,0,1,0) = (2,1,0)$$

I already determined, that the dimension of the kernel $$\varphi$$ will be $$2$$. What should be my next steps? I think I know how I would find such a matrix, that the dimension of the image is 2, but I don't know what to do about the second and the third point.

Thanks!

Simply define $$\;T:\Bbb R^4\to\Bbb R^3\;$$ by

$$T(1,1,1,1)=(1,2,1)\,,\,\,T(1,0,1,0):=(2,1,0)$$

and then complete $$\;\{v_1=(1,1,1,1)\,,\,\,v_2=(1,0,1,0)\}\;$$ to a basis of $$\;\Bbb R^4\;$$ and on the two other vectors define $$\;T\;$$ to be zero. For example, take $$\;\{v_3=(0,1,0,0)\,,\,\,v_4=(0,0,1,0)\}\;$$ . Check now that $$\;A:=\{v_1,v_2,v_3,v_4\}\;$$ is a basis of $$\;\Bbb R^4\;$$ , and (again), define:

$$Tv_1=(1,2,1)\,,\,Tv_2=(2,1,0)\,,\,\,Tv_3=Tv_4=(0,0,0)$$

and extend the above definition by linearity, thus obtaining a linear map. And now represent this map wrt the above basis in $$\;\Bbb R^4\;$$ and say the standard one $$\;B\;$$ in $$\;\Bbb R^3\;$$, getting :

$$[T]_A^B=\begin{pmatrix} 1&2&0&0\\ 2&1&0&0\\ 1&0&0&0\end{pmatrix}$$

• a typo: surely you mean $\mathbb{R^4}$ not $\mathbb{R}64$ Jun 9, 2019 at 21:47
• @peek-a-boo Of course. Thanks, edited. Jun 9, 2019 at 21:50
• and another one: in the last sentence you probably mean "the standard one $B$ in $\mathbb{R^3}$" not $A$ :) Jun 9, 2019 at 21:52
• Thanks...it's late over here. :) Jun 9, 2019 at 22:36
• Thanks. Could you please explain further how to represent the second map with the above basis in $\;\Bbb R^4\;$? Jun 10, 2019 at 11:58

Hint:

Let the matrix of $$\varphi$$ be the matrix $$A=\begin{pmatrix} a_1&b_1&c_1&d_1\\ a_2&b_2&c_2&d_2 \\ a_3&b_3&c_3&d_3 \end{pmatrix}.$$

The conditions on $$\varphi$$ can be written as $$\begin{cases} \varphi(e_1+e_2+e_3+e_4)=\scriptsize\begin{pmatrix} 1\\ 2 \\ 1 \end{pmatrix}\\[1ex] \varphi(e_1+e_3)=\scriptsize\begin{pmatrix} -1\\ 1 \\ 1 \end{pmatrix} \end{cases}\iff\begin{cases} \varphi(e_2+e_4)=\varphi(e_2)+\varphi(e_4)=\scriptsize\begin{pmatrix} 1\\ 2 \\ 1 \end{pmatrix}\\[1ex] \varphi(e_1+e_3)=\varphi(e_1)+\varphi(e_3)=\scriptsize\begin{pmatrix} 2\\ 1 \\ 0 \end{pmatrix} \end{cases}$$ So one obtains the linear systems $$\begin{cases} a_1+c_1=2,\\ a_2+c_2=1\\a_3+c_3=0 \end{cases},\qquad \begin{cases} b_1+d_1=-1,\\ b_2+d_2=1,\\b_3+d_3=1 \end{cases}$$ Can you proceed?

I can offer you a hint for an ad hoc solution for this exercise specifically (rather than all such questions in general).

One way to construct the matrix of such a linear transformation is to know the images under $$\varphi$$ of the standard basis vectors $$e_1=(1,0,0,0)$$, $$e_2=(0,1,0,0)$$, $$e_3=(0,0,1,0)$$, and $$e_4=(0,0,0,1)$$. Then you will put them down as the columns of the desired matrix: $$M = \begin{bmatrix} \varphi(e_1) & \varphi(e_2) & \varphi(e_3) & \varphi(e_4) \\ \end{bmatrix}.$$

Note that you already know $$\varphi(e_1)+\varphi(e_3)=\varphi(1,0,1,0)=(2,1,0)$$, as it's given to you. And you can also find $$\varphi(e_2)+\varphi(e_4)=\varphi(0,1,0,1)=\varphi(1,1,1,1)-\varphi(1,0,1,0)$$.

From here, you can make up as many examples satisfying the given conditions as you want. Pick any two vectors in $$\mathbb{R}^3$$ that add up to $$(2,1,0)$$ to be the values of $$\varphi(e_1)$$ and $$\varphi(e_3)$$. And pick any two vectors in $$\mathbb{R}^3$$ that add up to what you need to be the values of $$\varphi(e_2)$$ and $$\varphi(e_4)$$.

Let's pretend for a moment that points two and three require that:

• $$\phi(e_1) = w_1$$
• $$\phi(e_2) = w_2$$

Where $$e_i$$ are the elements of the canonical basis for $$\mathbb R^4$$, $$w_1 = (1,2,1)$$ and $$w_2 = (2,1,0)$$. An immediate solution for this problem would be the linear function defined by the associations above and sending $$e_3$$ and $$e_4$$ to the zero vector in $$\mathbb R^3$$, represented by the matrix:

$$A= \left(\begin{matrix}1&2&0&0\\2&1&0&0\\1&0&0&0\end{matrix}\right)$$

The kernel of this function has obviously dimension 2. Now to solve the original problem, let $$v_1=(1,1,1,1)$$ and $$v_2=(1,0,1,0)$$, and consider the change of basis matrix below:

$$B= \left(\begin{matrix}1&1&0&0\\1&0&0&0\\1&1&1&0\\1&0&0&1\end{matrix}\right)$$

You can see that $$Be_1 = v_1$$ and $$Be_2=v_2$$; so inverting the matrix we have $$B^{-1}v_1=e_1 \implies AB^{-1}v_1=w_1$$, and similarly $$AB^{-1}v_2=w_2$$. $$B^{-1}$$ is invertible, so $$\dim \ker AB^{-1} = \dim\ker A$$, hence the linear function represented by $$AB^{-1}$$ satisfies your requirements.

• To clarify if I understand correctly; when solving such problems in general, I can just complete both maps, create matrices and then do $AB^{-1}$? Jun 10, 2019 at 12:02
• @jamesF. Well yes. Of course, in this case I also had to verify that $w_1$ and $w_2$ were linearly independent: only then I could decide how many vectors had to be sent to $0$ by $A$. Jun 10, 2019 at 12:08
• Because if they were linearly dependent, it wouldn't span the entire space when adding two zero vectors? Jun 10, 2019 at 12:16
• $\phi$ shouldn't span the entire space: by the rank-nullity theorem, its span must have dimension 2. Jun 10, 2019 at 12:31

Find two vectors $$b_3,b_4$$ such that with $$b_1 = (1,1,1,1)^T, b_2 = (1,0,1,0)^T$$, the vectors $$b_1,...,b_4$$ span $$\mathbb{R}^4$$.

Note that $${\cal R} \phi = \operatorname{sp} \{ b_1,b_2\}$$, so you must have $$\phi(b_3), \phi(b_4) \in {\cal R} \phi$$, otherwise this would lead to a contradiction.

So you can choose $$\phi(b_3), \phi(b_4)$$ arbitrarily as long as they lie in $${\cal R} \phi$$. (Choosing the zero vector is an easy one.)

Once you have chosen these, this defines $$\phi$$ completely and the matrix representation $$A$$ is straightforward to obtain.

We have $$y_k = \phi(b_k) = A b_k = AB e_k$$, where $$B$$ is the matrix with columns $$b_1,...,b_4$$. From this we get $$Y=AB$$, where $$Y$$ is the matrix with columns $$y_1,...,y_4$$, and so $$A = Y B^{-1}$$.

• So when solving such problems in general, I just add basis vectors to both maps, so they span the entire space and calculate $YB^{-1}$? Jun 10, 2019 at 12:07
• It is hard to generalise, depends on what you mean by 'such problems'. Jun 10, 2019 at 14:00