# Finding a linear transformation given the null space

Find a linear transformation $$T\colon \mathbb{R^5} \to \mathbb{R^3}$$ such that $$\operatorname{null}(T) =\operatorname{span}\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}.$$

I tried to construct a matrix with columns corresponding to transformations of the basis vectors of $$\mathbb{R^5}$$ which are $$e_1=\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix},\quad e_2=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix},\quad e_3=\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\quad e_4=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\quad e_5=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}.$$

The matrix I made was $$\begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}.$$

I made this thinking that $$e_1$$ gets sent to $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$, $$e_2$$ gets sent to $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$, $$e_3$$ gets sent to $$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$, $$e_4$$ gets sent to $$\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$ and $$e_5$$ gets sent to $$\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$. Because we need $$T(e_3), T(e_4), T(e_5)$$ to be linearly independent. But this transformation I found out is just $$T(x_1,x_2,x_3,x_4,x_5) = (x_3,x_4,x_5)$$ and then it doesn't send $$e_1$$ or $$e_2$$ to $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$. I'm kind of confused. I want to find the linear transformation by constructing a matrix but it seems my work is wrong. How should I do this?

I also have another question. We have $$\operatorname{null}(T)=\operatorname{span}\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right\}.$$ How can I make an equation of the null space using these two basis vectors? I am asking this because I saw another question where we had to construct a linear transformation using the null space but the answer used the equation of the null space instead of the basis vectors of the null space. I, however, wish to do it both ways.

Thank you.

Yes, that matrix that you got corresponds to the linear map $$T(x_1,x_2,x_3,x_4,x_5)=(x_3,x_4,x_5)$$, but only if you consider $$\{e_1,e_2,e_3,e_4,e_5\}$$ as a basis of $$\Bbb R^5$$ (and the standard basis as a basis of $$\Bbb R^3$$). However, if you consider the standard basis as a basis of $$\Bbb R^5$$, then, since:

• the first vector of the standard basis is $$e_2-e_5$$,
• $$T(e_2-e_5)=(0,0,-1)^T$$;
• the second vector of the standard basis is $$e_1-e_2$$;
• $$T(e_1-e_2)=(0,0,0)^T$$,

the matrix of $$T$$ with respect to both standard bases is$$\begin{bmatrix}0&0&1&0&0\\0&0&0&1&0\\-1&0&0&0&1\end{bmatrix},$$and therefore $$T(x_1,x_2,x_3,x_4,x_5)=(x_3,x_4,-x_1+x_5)$$, and so indeed $$T(e_1)=T(e_2)=0$$.

• I don't quite understand. Why did we have to change the basis(e1,e2,e3,e3,e4,e5 (not the standard basis)) to the standard basis? Can't we work with the basis given? The basis e1,...,e5 that I wrote and then the standard basis of R^3. Nov 1, 2021 at 15:18
• I wish to know what I did wrong. How can we construct a matrix with respect to the basis (e1,..,e5) (that I wrote) and the standard basis of R^3. @José Carlos Santos Nov 1, 2021 at 15:29
• We can work with any bases whatsoever. What we cannot do is to look to the matrix that we got using a certain pair of bases and then to act as if it was the matrix of the linear map with respect to the standard bases, which is what you did when you “deduced” that $T(x_1,x_2,x_3,x_4,x_5)=T(x_3,x_4,x_5)$. Nov 1, 2021 at 15:33
• I don't understand, I wrote T($x_{1},x_{2},x_{3},x_{4},x_{5})$ = $(x_{3},x_{4},x_{5})$ and it is true, is it not? If we take any vector x = $(x_{1},x_{2},x_{3},x_{4},x_{5})$ in $\mathbb{R^5}$ and compute $Ax$, where A is the matrix I wrote, we get $(x_{3},x_{4},x_{5})$ Nov 1, 2021 at 15:42
• No, it is not true. It's only when the matrix with respect to the standard bases is equal to, say,$$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}$$that you can deduce that$$T(x,y,z)=(a_{11}x+a_{12}y+a_{13}z,a_{21}x+a_{22}y+a_{23}z).$$ Nov 1, 2021 at 15:48