# Find a basis $A$ of space $\mathbb R^{4}$ and basis $B$ of space $\mathbb R^{3}$

Let $$\varphi: \mathbb R^{4} \rightarrow \mathbb R^{3}$$: $$\varphi(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{3}+x_{4},x_{1}+x_{2}+2x_{3}+3x_{4},x_{1}-x_{2}-x_{4})$$Find a basis $$A$$ of space $$\mathbb R^{4}$$ and basis $$B$$ of space $$\mathbb R^{3}$$ such that $$M(\varphi)^{B}_{A}={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{bmatrix}}$$

I think that I know how to do this task. However I really need an assessment if it is correct.

My try:
$$M(\varphi)^{st}_{st}={\begin{bmatrix}1&0&1&1\\1&1&2&3\\1&-1&0&-1\end{bmatrix}}$$

After elementary operations on the matrix $$M(\varphi)^{st}_{st}$$ I have system of equations: $$\begin{cases} x_{1}+x_{3}+x_{4}=0 \\ x_{2}+x_{3}+2x_{4}=0\end{cases}$$

So $$\ker \varphi=lin\left\{(-1,-1,1,0),(-1,-2,0,1)\right\}$$. This vectors are in a basis $$A$$ but I need two more vectors.

$$\dim(\ker\varphi)=2$$ so $$\dim(im \varphi)=2$$ and it can be: $$(1,1,1),(0,1,-1)$$ (I take $$2$$ linearly independent vectors from the matrix $$M(\varphi)^{st}_{st}$$ columns). This vectors are in a basis $$B$$.

In this moment I have $$A=\left\{\alpha_{1}, \alpha_{2},(-1,-1,1,0),(-1,-2,0,1)\right\}$$ and $$B=\left\{(1,1,1),(0,1,-1),\beta_{3}\right\}$$

From $$M(\varphi)^{B}_{A}$$ I have:
$$\varphi(\alpha_{1})=1 \cdot \beta_{1}$$
$$\varphi(\alpha_{2})=1 \cdot \beta_{2}$$
$$\varphi(\alpha_{3})=0$$
$$\varphi(\alpha_{4})=0$$

That is why:

$$\varphi(\alpha_{1})=(1,1,1)$$ and $$\alpha_{1}=(1,0,0,0)$$
$$\varphi(\alpha_{2})=(0,1,-1)$$ and$$\alpha_{2}=(0,1,0,0)$$
$$\beta_{3}$$ I can choose anyway because I do not have any dependencies on it. It can be linerly indipendent from other basis vectors, so it can be $$\beta=(0,0,1)$$

I will be extremely grateful for checking this solution and indicating any errors.

$$\beta_{3}$$ I can choose anyway because I do not have any dependencies on it. It can be linerly indipendent from other basis vectors, so it can be $$\beta=(0,0,1)$$

Perhaps you mean it well, but you don't want "$$\beta_{3}$$ (...) can be linearly independent (...)" but rather has to be because otherwise, $$B$$ wouldn't be a basis. So you are right that you can freely choose $$\beta_{3}$$, as long as it's linearly independent from the first two elements of $$B$$.

So $$\ker \varphi=lin\left\{(-1,-1,1,0),(-1,-2,0,1)\right\}$$. This vectors are in a basis $$A$$ but I need two more vectors.

$$\dim(\ker\varphi)=2$$ so $$\dim(im \varphi)=2$$ and it can be: $$(1,1,1),(0,1,-1)$$ (I take $$2$$ linearly independent vectors from the matrix $$M(\varphi)^{st}_{st}$$ columns)

I would do this differently, but perhaps because I'm not sure if I follow your reasoning.

Based on the kernel, you already have the last two (blue) elements in $$A=\left\{\alpha_1,\alpha_2,\color{blue}{\alpha_3},\color{blue}{\alpha_4}\right\}$$, so now you can simply extend to a full basis of $$\mathbb{R^4}$$ by picking any $$\alpha_1$$ and $$\alpha_2$$, as long as all four are linearly independent. The first two standard basis vectors work, so pick e.g. $$\alpha_1=(1,0,0,0)$$ and $$\alpha_2=(0,1,0,0)$$. The required form of the matrix is then automatically satisfied if you pick $$\beta_1 = \varphi\left(\alpha_1\right)$$ and $$\beta_2 = \varphi\left(\alpha_2\right)$$ and then add $$\beta_3$$ as described above.

I have the feeling your approach is the other way around: finding suitable $$\alpha_1$$ and $$\alpha_2$$ for $$A$$ to match earlier picked $$\beta_1$$ and $$\beta_2$$ in $$B$$.

Thumbs up for the well documented question, showing your own work.

• Regarding the linear independence of $\beta_ {3}$, you're right that it should be there "has to be", this is just a blunder from my side but thank you for being vigilant.Regarding the further part of your speech: so I have to understand that with $M (\varphi)^ {st}_{st}$ I just need to use only $ker \varphi$, then select linearly independent vectors to unbutton the basis $A$ and match the vectors with $B$ so that the matrix $M (\varphi)^ {B}_{A}$ that it would be true for the designated basis $A$. Did I understand everything correctly? – MP3129 Mar 13 at 17:57
• I think you have it yes: you start with a basis for the kernel and take those as last vectors of basis $A$ (this creates the desired zero-columns at the end of the matrix), you then complete this 'partial basis' to a basis for $A$. If you then take the images of these added vectors as base vectors in $B$ (same order), you create the desired 'unity columns' at the start of the matrix. You only have to complete the basis $B$ but this does not affect the matrix. – StackTD Mar 14 at 8:58
• Ok, thank you very much for getting to know my understanding and such an extensive indication of all errors, because first and foremost I wanted to check my way – MP3129 Mar 14 at 20:29

You are guessing. There is no need to.

If the required bases are, respectively $$\{v_1,v_2,v_3,v_4\}$$ and $$\{w_1,w_2,w_3\}$$, then the condition for the matrix translates into $$\varphi(v_1)=w_1,\quad \varphi(v_2)=w_2,\quad \varphi(v_3)=0,\quad \varphi(v_4)=0$$ The idea is then to complete a basis of the kernel, which you performed correctly.

Since $$\ker\phi$$ is the set of vectors satisfying $$\begin{cases} x_{1}+x_{3}+x_{4}=0 \\ x_{1}+x_{2}+2x_{3}+3x_{4}=0\\ x_{1}-x_{2}-x_{4}=0 \end{cases}$$ you can do elimination $$\begin{bmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 2 & 3 \\ 1 & -1 & 0 & -1 \end{bmatrix}\to \begin{bmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & -1 & -1 & -2 \end{bmatrix}\to \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ so a basis of $$\ker\varphi$$ is given by $$v_3=(-1,-1,1,0)$$, $$v_4=(-1,-2,0,1)$$. Now complete to a basis of $$\mathbb{R}^4$$ finding the null space of $$\begin{bmatrix} -1 & -1 & 1 & 0 \\ -1 & -2 & 0 & 1 \end{bmatrix}$$ The elimination yields $$\to \begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & -1 & -1 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & 1 & 1 & -1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & -1 \end{bmatrix}$$ which provides $$v_1=(2,-1,1,0)$$ and $$v_2=(-1,1,0,1)$$. Note that, by construction, $$\{\varphi(v_1),\varphi(v_2)\}$$ is linearly independent. Now take $$w_1=\varphi(v_1)=(3,3,2)\qquad w_2=\varphi(v_2)=(0,3,-3)$$ and complete this to a basis of $$\mathbb{R}^3$$ finding the null space of $$\begin{bmatrix} 3 & 3 & 2 \\ 0 & 3 & -3 \end{bmatrix}$$ Elimination: $$\to \begin{bmatrix} 1 & 1 & 2/3 \\ 0 & 1 & -1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 5/3 \\ 0 & 1 & -1 \end{bmatrix}$$ so the vector you need is $$w_3=(-5/3,1,1)$$.