Find a subset of $S \subseteq N(T)$ that is a basis of the Null of $T$.

Question

I'm given that $T$ is the linear transformation $T:\mathbb{R}^5 \rightarrow \mathbb{R}^2$ defined by $T(x_1,x_2,x_3,x_4,x_5)=(x_1-2x_2+x_5,x_1+x_3)$.

I need to (if possible) find a subset of $S \subset N(T)$ that's a basis of the Null of $T$.

One of the given sets S that I need to solve for is $S = \{(2,1,-2,0,0),(0,1,0,0,2) \}.$

I don't recall going over this in class yet, and I can't find an example in my book, so I don't know how to go about this problem.

I'm guessing that I need to set up a matrix using the vectors given by $T$ and $S$. I just don't know what I'm supposed to do after that.

Any help is much appreciated.

Answer 1

It's always possible to find a subset of the null space of a linear transformation which forms a basis. This is a method that will work in general: $T$ corresponds to multiplication by the matrix $\begin{pmatrix}1&-2&0&0&1\\1&0&1&0&0\end{pmatrix}$.

Now row-reduce: $\to\begin{pmatrix}1&-2&0&0&1\\0&2&1&0&-1\end{pmatrix}\to\begin{pmatrix}1&-2&0&0&1\\0&1&1/2&0&-1/2\end{pmatrix}\to\begin{pmatrix}1&0&1&0&0\\0&1&1/2&0&-1\end{pmatrix}$.

At this point we "back-substitute": $x_2+1/2x_3-x_5=0\implies x_2=-1/2x_3+x_5$. And from the first row: $x_1+x_3=0\implies x_1=-x_3$.

So we have that $N(t)=\{(-x_3,-1/2x_3+x_5,x_3,x_4,x_5): x_3,x_4,x_5\in \Bbb R\}=\{x_3(-1,-1/2,1,0,0)+x_4(0,0,0,1,0)+x_5(0,1,0,0,1):x_3,x_4,x_5\in\Bbb R\}=\rm{span}\{(-1,-1/2,1,0,0),(0,0,0,1,0),(0,1,0,0,1)\}$.

Since the rank of the matrix is $2$, the nullity is $3$. Thus we have arrived at a basis.

Answer 2

It's not necessary to set up a matrix. Notice that $T$ is surjective, because for any $(x, y) \in \mathbb{R}^{2}$ we have $T(0, 0, y, 0, x) = (x , y)$. Thus,

$$ \dim \ker T = \dim \mathbb{R}^{5} - \dim \operatorname{im} T = \dim \mathbb{R}^{5} - \dim \mathbb{R}^{2} = 3. $$

This means that we only need to find three linearly independent vectors in $\mathbb{R}^{5}$ that become $(0, 0)$ under $T$ and we'll have a basis for $\ker T$. By simple inspection, we see that $(2, 1, -2, 0, 0)$ and $(0,1, 0, 0, 2)$ become $(0, 0)$ under $T$ and are linearly independent, so we only need one more vector, say $(0, 0, 0, 1, 0)$. This vector also becomes $(0, 0)$ under $T$ and is independent from the previous vectors (the $1$ in the fourth entry does the trick) , so the list $$ (2, 1, -2, 0, 0), (0, 1, 0, 0, 2), (0, 0, 0, 1, 0) $$ is a basis for $\ker T$.