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 $T_2 \circ T_1$

Question

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?

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(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

$$\mathbf{M}_1 = \begin{pmatrix}1&4&-5&8\\0&-4&1&-5\\-1&-3&0&-2\\0&1&1&0\end{pmatrix}\text{ and }\mathbf{M}_2 = \begin{pmatrix}1&2&-1&3\\1&0&4&5\\3&2&7&13\\1&4&-6&1\end{pmatrix}$$

1. Show that the rank of $\mathbf{M}_1$ is equal to 3.
2. Write down a basis for $R_1$, the range space of $T_1$, and find a basis for the null space of $T_1$.
3. Find a basis for $K_2$, the null space of $T_2$, and hence show that $K_2$ is a subspace of $R_1$.
4. Hence, or otherwise, find three linearly independent vectors in the null space of $T_3$.

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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)
• ???

Answer 1

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.

Answer 2

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$(T_2)$.