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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.

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    $\begingroup$ elements ($x_1,x_2,x_3$) of kernel have $x_2+x_3=2x_2+2x_3=0$ $\endgroup$ – J. W. Tanner Oct 29 '19 at 12:39
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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 $ker(T).$

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    $\begingroup$ 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? $\endgroup$ – James Debenham Oct 29 '19 at 12:56
  • $\begingroup$ Oooops, a typo, yes delete the 4-th entries. $\endgroup$ – Fred Oct 29 '19 at 13:03
  • $\begingroup$ How did you figure that the rank(T) = 1? Is that because there is only one linearly independent vector in the matrix A? $\endgroup$ – James Debenham Oct 29 '19 at 13:54
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You're wrong. This matrix has to be multiplied by a $3{\times}1$ column vector. So there's a single equation: $\;x_2+x_3=0$.

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

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