If $AB=0$, how can I show that $Col(B)$ is a subspace of $Nul(A)$?

Question

Let $A$ and $B$ be matrices such that $AB = 0$. Show that $Col(B)$ is a subspace of $Nul(A)$

Till now i have that $Col(B)$ is the vector $b$, such that $Bx=b$, and from this i get: $$AB = 0 \Rightarrow ABx = 0x = 0 \Rightarrow ACol(B) = 0$$

I assume that this has something to do with $Nul(A)$ being the set of all x that satisfies the equation $Ax=0$, but i am not quite sure how to link these up to the def. of a subspace.

Answer 1

You can prove that $X$ is a subspace of $Y$ by proving

1. That $X$ is a subset of $Y$
2. That $X$ is a vector space.

---

In your case, you need to prove that $col(B)$ is a vector space, which should be fairly obvious.

Additionally, you need to prove that $col(B)$ is a subset of $Nul(A)$, which can be proven like in general like so:

• Take any $b\in Col(B)$.
• Therefore $somethingsomething$
• Therefore, $Ab = 0$
• Therefore, $b\in Nul(A)$
• Therefore, $Col(B)\subseteq Nul(A)$

Now, give it a try and let us know how far you got.

---

Hint:

• Let $b_1,b_2\dots$ be the columns of $B$. How are $b_i$ connected to $b$?
• What is $Ab_i$ equal to?
• What is $A(\alpha_1 b_1 + \cdots \alpha_n b_n)$ equal to?

Answer 2

Thank you so much! This is what I've got so far:

$$Ab_i = ABx_i = 0 $$ Therefore all $$ b_i $$ is in the null space.

$$ A(\alpha_1b_1+...+\alpha_nb_n) = ACol(B)\qquad(?)$$