# Finding the kernel and basis for the kernel of a linear transformation

For the linear mapping: $$T: \mathbb{R}^3 \to \mathbb{R}^4: (x_1, x_2, x_3) \mapsto (0 , x_2 + x_3, 0, 2x_2 + 2x_3).$$

I've been asked to find the kernel, the basis for the kernel and hence the nullity $$n(T)$$.

So far, I've established the matrix $$A$$ to represent the linear map after applying the transformation $$T$$ to the standard basis of $$\mathbb{R}^3$$ like such:

$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 2 & 2 \\ \end{bmatrix}$$

I believe this is correct so far. Then, to find the kernel I've set up the equation below, since $$\ker(A)$$ is the set of vectors for which when the transformation is applied, it equals zero.

$$x_2 + x_3 = 0$$ (now edited to be correct)

Where would I go from here? Thanks.

• elements ($x_1,x_2,x_3$) of kernel have $x_2+x_3=2x_2+2x_3=0$ – J. W. Tanner Oct 29 '19 at 12:39

The matrix $$A$$ is correct. We have $$rank(A)=1$$, hence, by the nullity - rank - theorem we get $$n(T)=2.$$ It is easy to see that $$a:=(1,0,0)$$ and $$b:=(0,1,-1)$$ are elements of $$ker(T).$$ Since $$a$$ and $$b$$ are linearily independent, we get that $$\{a,b\}$$ is a basis of $$ker(T).$$

• Thank you for your answer. The other answer stated that the kernel is a plane in $\mathbb{R}^3$, but your answer seems to suggest that the kernel has 4 dimensions. Where is this confusion coming from? Is the kernel the same dimension as the source space or the target space? – James Debenham Oct 29 '19 at 12:56
• Oooops, a typo, yes delete the 4-th entries. – Fred Oct 29 '19 at 13:03
• How did you figure that the rank(T) = 1? Is that because there is only one linearly independent vector in the matrix A? – James Debenham Oct 29 '19 at 13:54

You're wrong. This matrix has to be multiplied by a $$3{\times}1$$ column vector. So there's a single equation: $$\;x_2+x_3=0$$.

• Of course! Pretty rookie error there. Thank you. – James Debenham Oct 29 '19 at 12:44
• does this mean that $x_1$ is set to be arbitrary in the set of solutions? – James Debenham Oct 29 '19 at 12:46
• Yes. The kernel is a plane in $\mathbf R^3$. B.t.w., you don't need the matrix to see that. – Bernard Oct 29 '19 at 12:48
• Can you explain why I don't need the matrix to see that? And further to this question, how would I find the basis for that set of solutions? – James Debenham Oct 29 '19 at 12:49
• It already is in the defining equations for $T$. – Bernard Oct 29 '19 at 12:51