# Dimension of Nullspace resulting from two linear transformations

Given linear transformations $T_1 = M_1$ and $T_2 = M_2$, the linear transformation $T_3$ is the result of applying $T_1$ and $T_2$ in that order.

I've sort of adapted this question from some homework, so here are some extra details if required, $T_1 , T_2 , T_3$ all transform $\Bbb{R}^4 \rightarrow \Bbb{R}^4$, $dim(basis\;T_1)=3$, $dim(ker\;T_2)=2$, and the nullspace of $T_2$ is a subspace of the basis of $T_1$.

How do I find the nullspace of $T_3$, without multiplying the matrices $M_2$ and $M_1$ and then solving the nullspace of the resultant matrix?

Although solving the resultant nullspace of the $M_2M_1$ matrix will give me the answer, I would like to know if there are other/ better ways to solve such a problem.

In general for $T_1:U\to V$ and $T_2:V\to W$, you can write
$U=\ker T_1\oplus r(T_1)$ where a basis for $r(T_1)$ is $u_1,...,u_k$ with $T_1(u_1),...,T_1(u_k)$ some basis for the image of $T_1$. Similarly for $V=\ker T_2\oplus r(T_2)$.
To see the kernel of the composition, note that it will always contain the $\ker T_1$ as well as those elements of $\ker T_2$ that are images of vectors under $T_1$, that is:
$\ker(T_2T_1)=\ker T_1\oplus r(im(T_1)\cap \ker(T_2))$.
In your case, $U=V$ and $im(T_1)\supset \ker (T_2)$, so $\ker(T_2T_1)=\ker T_1\oplus r(\ker T_2)$.