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.
Best Answer
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 $ker(T).$