Let $A$ be an $m \times n$ matrix and $B$ be an $n \times p$ matrix. Let

$$S=\left \{Bx\mid x\in \mathbb{R}^p \text{ and } ABx=0\right \}$$

Prove that $\dim(S)=\operatorname{rank}(B)-\operatorname{rank}(AB)$

What I thought is actually the collection of vector spanning by the column vector of $B$, so it seems that $\operatorname{rank}(S)$ is somehow equal to $\operatorname{rank}(B)$ (I don't know how to prove it, or maybe I'm wrong)
Then, I tried to use rank-nullity theorem
$$\dim(S)=\operatorname{rank}(B)+\operatorname{rank}(S)$$ But I already get stuck here

  • $\begingroup$ It is helpful to take a vector space and linear transformation point of view. $\endgroup$ – Zhanxiong Jun 27 '15 at 17:45

Identify $A \in L(\mathbb{R}^n, \mathbb{R}^m)$, $B \in L(\mathbb{R}^p, \mathbb{R}^n)$, and $AB \in L(\mathbb{R}^p, \mathbb{R}^m)$. Let $\mathscr{R}(A), \mathscr{R}(B), \mathscr{R}(AB)$ be the range subspaces of $A$, $B$, $AB$, respectively, and let $\mathscr{N}(A), \mathscr{N}(B), \mathscr{N}(AB)$ be the null spaces of $A$, $B$, $AB$, respectively.

By definition and the rank-nullity theorem: \begin{align} & \text{rank}(B) = \dim(\mathscr{R}(B)) = p - \dim(\mathscr{N}(B)) \tag{1} \\ & \text{rank}(AB) = \dim(\mathscr{R}(AB))= p - \dim(\mathscr{N}(AB)) \tag{2} \end{align}

Notice that $S$ can be viewed as the range space of the linear transformation $B$ restricted to the subspace $\mathscr{N}(AB)$, apply the rank-nullity theorem once more, it follows that $$\dim(S) + \dim(\mathscr{N}(B)) = \dim(\mathscr{N}(AB))$$

Plug $(1)$ and $(2)$ into the above equation, we conclude

$$\dim(S) + (p - \text{rank}(B)) = p - \text{rank}(AB)$$

That is, $\dim(S) = \text{rank}(B) - \text{rank}(AB)$.


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.