I understand this proof of a related fact: Show that $nullity(B)\leq nullity(AB)$ but I'm not sure if it translates to this due to matrix dimensions. If $A$ and $B$ are square, we can say:

Let $x$ be an element of the null space of $A$. Then $(AB)x=(Ax)B=(0)B=0$ so $x$ is an element of the null space of $AB$, implying that $NS(A)\subseteq NS(AB)$, and hence the desired result. But since the matrices aren't necessarily square, can we adapt this proof using the left null space? ie...

Let $x^T$ be an element of the left null space of $A$. Then $x^T(AB)=(x^TA)B=(0)B=0$, so $x^T$ is an element of the left null space of $AB$, and hence nullity$(A)\leq$ nullity$(AB)$.

I'm not sure if this works because I'm not entirely clear on the relation between the left null space and the null space.

  • $\begingroup$ Ah, careless mistake. So this doesn't work even if they are square matrices. $\endgroup$ – Atsina Jan 3 '18 at 0:38
  • $\begingroup$ Your equality $(AB)x=(Ax)B)$ is false, because you can′t multiply the matrix $B$ on the left by the column-vector $Ax$. $\endgroup$ – Bernard Jan 3 '18 at 0:42

$A$ has smaller left-nullity than $AB$ but not smaller (right)-nullity. For example, if

$$ A = \begin{pmatrix} 1 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, $$

then $\operatorname{nullity}(A) = 1$ but $\operatorname{nullity}(AB) = 0$.

You can see what the rank-nullity theorem says: if the dimension of $A$ is $m \times n$ and $B$ is $n \times p$ then

$$\operatorname{rank}(A) + \operatorname{nullity}(A) = n,$$

$$\operatorname{rank}(B) + \operatorname{nullity}(B) = p,$$

$$\operatorname{rank}(AB) + \operatorname{nullity}(AB) = p.$$

So we are able to relate the nullity of $B$ to that of $AB$ because they are both related to $p$ but the nullity of $A$ is only related to $n$. If $\operatorname{nullity}(A) > p$ then $\operatorname{nullity}(A) > \operatorname{nullity}(AB)$ since $\operatorname{nullity}(AB) \le p$.

If $A$ and $B$ are square matrices then the inequality is always true: $\operatorname{nullity}(A) \le \operatorname{nullity}(AB)$. One way to see this is to observe that the rank decreases: $\operatorname{rank}(AB) \le \operatorname{rank}(A)$.


For square matrices, assuming you already know $\operatorname{nullity}(B) \le \operatorname{nullity}(AB)$, you can prove $\operatorname{nullity}(A) \le \operatorname{nullity}(AB)$ by taking the transpose.

The nullity of the transposed matrix is equal to the nullity of the original one so:

$$\operatorname{nullity}(A) = \operatorname{nullity}\left(A^T\right) \le \operatorname{nullity}\left(B^TA^T\right) = \operatorname{nullity}\left((AB)^T\right) = \operatorname{nullity}\left(AB\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.