Null space of matrix product Is the following relation true or untrue?
$$\ker(ABC)=\ker(B)$$
where $A,B,C\in\mathbb R^{n\times n}$, $A$ and $C$ are nonsingular, and $B$ is singular.
 A: Not true. Take for example $$B=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$$
which has a null space spanned by the vector $[0,1]^T$ and take $$A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\\
C = \begin{bmatrix}0&1\\1&0\end{bmatrix}$$
The product $$ABC = \begin{bmatrix}0&1\\0&0\end{bmatrix}$$ has a null space spanned by the vector $[0,1]^T.$
A: Not true in general as the other answer shows. If $C$ is nonsingular and $Z$ is a basis$^{\color{red}*}$ of the kernel of $B$, then $C^{-1}Z$ is a basis of the kernel of $BC$ and hence of $ABC$ provided $A$ is nonsingular as well. So unless $C^{-1}Z=Z$ or equivalent $Z=CZ$, that is, $C$ acts as the identity on $\mathrm{ker}(B)$, one has $\mathrm{ker}(ABC)\neq\mathrm{ker}(B)$.
$^{\color{red}*}$$Z=[z_1,\ldots,z_k]$, $k=\dim\mathrm{ker}(B)$, $\mathcal{R}(Z)=\mathrm{span}\{z_1,\ldots,z_k\}=\mathrm{ker}(B)$. 
A: The general statement here is about linear maps between $4$ vector spaces:
$$
V_1\overset{\varphi_1}\longrightarrow V_2
\overset{\varphi_2}\longrightarrow V_3
\overset{\varphi_3}\longrightarrow V_4,
$$
where $\varphi_1$ and $\varphi_3$ are isomorphisms (corresponding to the non-singular matrices) and you want to know about $\ker(\varphi_3\circ\varphi_2\circ\varphi_1)$.

Claim: $\ker(\varphi_3\circ\varphi_2\circ\varphi_1) = \ker(\varphi_2\circ\varphi_1) = \varphi_1^{-1}(\ker \varphi_2).$

Proof. Let $v \in \ker(\varphi_3\circ\varphi_2\circ\varphi_1)$, that is $\varphi_3(\varphi_2(\varphi_1(v)))=0$. Since $\varphi_3$ is an isomorphism, this is equivalent to $\varphi_2(\varphi_1(v))=0$, i.e. $v\in\ker(\varphi_2\circ\varphi_1)$, which proves the first equality. For the second equality, whenever $\varphi_2(\varphi_1(v))=0$ we have $\varphi_1(v)\in\ker \varphi_2$, so $v\in\varphi_1^{-1}(\ker\varphi_2)$. On the other hand, whenever $v\in\varphi_1^{-1}(\ker\varphi_2)$ we have $\varphi_1(v)\in \ker\varphi_2$, so $\varphi_2(\varphi_1(v))=0$, i.e. $v\in\ker(\varphi_2\circ\varphi_1)$.
Translated to the finite-dimensional case represented by matrices $A,B,C$ this says
$$
\ker(ABC) = \ker(BC) = C^{-1} \ker B.
$$
