3
$\begingroup$

Let $A \in \mathbb{C}^{m\times n}$, and $B \in \mathbb{C}^{n\times k}$ complex matrices.

If A is full-column rank matrix then can we say that rank($AB$) = rank ($BA$) = rank($B$)?

What can we say about $N(AB)$?

Under what condition $N(AB) = N(B)$?

I need help to understand this. I would be very much helpful for any kind of help and suggestions.

$\endgroup$
  • $\begingroup$ Suggest you compute some examples. $\endgroup$ – Gerry Myerson Apr 8 '14 at 12:56
  • 2
    $\begingroup$ $BA$ is only defined for $m=k$ $\endgroup$ – daw Apr 8 '14 at 17:16
4
$\begingroup$

Supposing $m=k$ so that $BA$ is defined, we can use Sylvester's rank inequality combined with

rank$(AB)\leq$ min$\{$rank$(A)$, rank$(B)\}$. So we have

rank$(A)+$rank$(B)-n \leq$ rank$(AB)\leq$ min$\{$rank$(A)$, rank$(B)\}.$

If $A$ is full column rank then we must have rank$(B) \leq$ rank$(A)$, and we have rank$(A)=n$, so the inequality reduces to

rank$(B) \leq$ rank$(AB)\leq$ rank$(B),$

and we have a similar result for rank$(BA)$, proving the first part.

Note that $N(B) \subseteq N(AB)$ - it is easy to prove this using the standard way of taking $v \in N(B)$ and then showing that we must have $v \in N(AB)$.

You will have the last equality if $A^{-1}$ exists...again this is not difficult to prove - you must just show that $v \in N(AB)$ implies $v \in N(B)$ if $A$ is invertible.

$\endgroup$
  • 2
    $\begingroup$ One does not have a similar result for rank$(BA)$. $\endgroup$ – Marc van Leeuwen Dec 3 '14 at 9:41
2
$\begingroup$

Summary: the equation $\def\rk{\operatorname{rank}}\rk(AB)=\rk(B)$ holds, but (supposing $BA$ is defined) one can have $\rk(BA)\neq\rk(B)$. Simple example of the latter: $$ A=\begin{pmatrix}1\\0\end{pmatrix} ,\qquad B=\begin{pmatrix}0&1\end{pmatrix} ,\qquad BA=(0) ,\qquad\text{( while } AB=\begin{pmatrix}0&1\\0&0\end{pmatrix} \text{).} $$ One does have (with the given hypotheses) $\ker(AB)=\ker(B)$, which more generally holds whenever $\ker(A)\cap\operatorname{range}(B)=\{0\}$ (the given hypotheses say that $\ker(A)=\{0\}$, which is stronger).


Saying that $A$ has full column rank is saying that the columns of $A$ are linearly independent, which means that the map $x\mapsto Ax$ is injective: since $Ax$ is forming a linear combination of the columns of $A$ with the entries of $x$ as coefficients, linear independence says that $Ax=0$ only has the trivial solution $x=0$. Alternative formulation: $\ker(A)=\{0\}$.

Now an injective linear map sends linearly independent sets to linearly independent sets, so a maximal independent set of columns of $B$ corresponds (at the same column indices) to a maximal independent set of columns of$~AB$, which gives $\rk(AB)=\rk(B)$. As the counterexample shows this argumentation does not work when $A$ is the right factor, as in $BA$.

As for $\ker(AB)$, clearly every $x$ with $Bx=0$ also has $ABx=0$, so $\ker(B)\subseteq\ker(AB)$ without any hypotheses on $A$ (other than that $AB$ is defined). For a condition for the reverse inclusion; if we imagine that $Bx\neq0$ but nonetheless $ABx=0$, this means that $Bx$ is a nonzero vector in $\ker(A)$. Clearly when $\ker(A)=\{0\}$ this cannot happen, and more generally it cannot whenever $\ker(A)\cap\operatorname{range}(B)=\{0\}$, since $Bx$ must be a vector of that intersection.

$\endgroup$
  • $\begingroup$ I think you meant $\textrm{rank}(AB) = \textrm{rank}(B)$, not $\textrm{rank}(AB) = \textrm{rank}(A)$ in your proof. $\endgroup$ – Glassjawed Oct 19 '15 at 21:30
  • $\begingroup$ @ClarkKent: Right; I corrected it now. Thank you. $\endgroup$ – Marc van Leeuwen Oct 20 '15 at 4:32
0
$\begingroup$

$rank(AB)≤ \min\{rank(A), rank(B)\}$ implies $rank(AB) ≤ rank(B)$

Also,Full rank matrices are invertible implies A is invertible.

$rank(B)=rank(A^{−1}AB)≤ \min\{rank(A^{−1}), rank(AB)\},A^{−1}$ means the inverse of A. i.e, $rank(B)≤rank(AB)$

Thus, rank $(AB)=rank(BA)=rank(B)$.

Always, $N(B)⊆N(AB)$,since $B(x)=0$ implies $AB(x)=0$. Converse is true when A is invertible.

Hope you understood!

$\endgroup$

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.