If the null space of $T_2$ is a subspace of the range of $T_1$, deduce three linearly independent vectors in the null space of $T_2 \circ T_1$ Below is a question from an old A Level Further Mathematics paper. Parts 1 through 3 are straightforward, but part 4 has stumped me. It is easy to go the long way and just find the basis for the null space, but what is the hence way?

(Further Mathematics Paper 1, Nov 1997)
The linear transformations $T_1:\mathbb{R}^4\rightarrow\mathbb{R}^4$, $T_2:\mathbb{R}^4\rightarrow\mathbb{R}^4$, $T_3:\mathbb{R}^4\rightarrow\mathbb{R}^4$ are represented by the matrices $\mathbf{M}_1$, $\mathbf{M}_2$ and $\mathbf{M}_2\mathbf{M}_1$ respectively, where
$$\mathbf{M}_1 = \begin{pmatrix}1&4&-5&8\\0&-4&1&-5\\-1&-3&0&-2\\0&1&1&0\end{pmatrix}\text{ and }\mathbf{M}_2 = \begin{pmatrix}1&2&-1&3\\1&0&4&5\\3&2&7&13\\1&4&-6&1\end{pmatrix}$$


*

*Show that the rank of $\mathbf{M}_1$ is equal to 3.

*Write down a basis for $R_1$, the range space of $T_1$, and find a basis for the null space of $T_1$.

*Find a basis for $K_2$, the null space of $T_2$, and hence show that $K_2$ is a subspace of $R_1$.

*Hence, or otherwise, find three linearly independent vectors in the null space of $T_3$.



Answers to part 1–3
Basis for $R_1 = \left\{\begin{pmatrix}1\\0\\-1\\0\end{pmatrix},\begin{pmatrix}4\\-4\\-3\\1\end{pmatrix},\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}\right\}$
Basis for the null space of $T_1 = \left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}\right\}$
Basis for $K_2 = \left\{\begin{pmatrix}-8\\5\\2\\0\end{pmatrix},\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}\right\}$
Part 4
Three linearly independent vectors in the null space of $T_3$ are


*

*$\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}$ (from the basis for the null space of $T_1$)

*$\begin{pmatrix}4\\-4\\-3\\1\end{pmatrix}$ (this is just a coincidence, as far as I can tell)

*???

 A: Note that Ker$(T_2)+$Ker$(T_1)$ is a subspace of Ker$(T_2\circ T_1)$. Additionaly, in this case their intersection is $0$ so you can use already computed $3$ vectors in Ker$(T_1)$ and Ker$(T_2)$.
A: First note that $\ker T_1 \subseteq \ker T_2T_1$ so you already have one vector, the one in $\ker T_1$.
$$\ker T_1 = \operatorname{span}\left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}\right\}, \quad\ker T_2 = \operatorname{span}\left\{\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}, \begin{pmatrix}8\\-5\\-2\\0\end{pmatrix}\right\}$$
Now notice that the basis vectors for $\ker T_2$ are in fact the third and the fourth columns of the matrix $M_1$, respectively. Therefore
$$\begin{pmatrix}-5\\1\\0\\1\end{pmatrix} = T_1e_3 = T_1\begin{pmatrix}0\\0\\1\\0\end{pmatrix}$$
$$\begin{pmatrix}8\\-5\\-2\\0\end{pmatrix} = T_1e_4 = T_1\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$
So
$$\left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}$$
are three vectors in $\ker T_2T_1$ which are clearly linearly independent.
