Dimension of image of transformation

Question

T:$\Bbb{R}^3\to\Bbb{R}^{2\times2}$ is a linear transformation such that

$$T(\begin{bmatrix}x & y & z\end{bmatrix})=\begin{bmatrix} y & x-z \\ 4x-4z & 0 \end{bmatrix} $$

find the $\operatorname{Ker}(T)$ and the dimension of image of $T$.

I have shown that it is a linear transformation and found $\operatorname{Ker}(T)$ , but need help on dimension of image. Thanks

Answer 1

Check what $T$ does the the basis vectors:

$$T(\begin{bmatrix}1 & 0 & 0\end{bmatrix})=\begin{bmatrix} 0 & 1 \\ 4 & 0 \end{bmatrix} $$

$$T(\begin{bmatrix}0 & 1 & 0\end{bmatrix})=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$
$$T(\begin{bmatrix}0 & 0 & 1\end{bmatrix})=\begin{bmatrix} 0 & -1 \\ -4 & 0 \end{bmatrix} $$

See then that $T([1,0,1])=\vec0.$ Notice that the first first two computations yield linearly independent vectors. So the rank is 2. The nullity is 1, since the kernel of this map is spanned by $[1,0,1].$ You can verify this using the rank nullity theorem.

Answer 2

Guide:

Use rank-nullity theorem, we have

$$\operatorname{dim}(\operatorname{Ker}(T)) + \operatorname{rank}(T)= 3$$