Proving that $rank(A) + rank(B) - n \le rank(AB)$

Question

I have to demonstrate the following statement:

If $\ A \in M_{m,n}(\mathbb{R})$ and $\ B \in M_{n,t}(\mathbb{R})$ show that: $$ rank(A)+ rank(B) -n \ \le \ rank(AB) \ \le \ \min\{rank(A), rank(B)\} $$

So far I've only managed to show the second inequality (I think), because $rank(A) \ge rank(AB)$ and $rank(B) \ge rank(AB)$. I don't know how to approach the first one. It could probably be interpreted using dimensions of images and matrices of linear applications, but then I can't figure out what $n$ corresponds to.

Any help/hints would be appreciated.

score 2 · Accepted Answer · answered Jan 22 '23 at 13:35

2

$$ rank(A) + rank(B) \le rank\begin{pmatrix} A & 0 \\ -I_n & B \end{pmatrix} = rank\begin{pmatrix}0 & AB \\ -I_n & 0 \end{pmatrix} = rank(AB) + n $$

Can you find the transform that turns $\begin{pmatrix} A & 0 \\ -I_n & B \end{pmatrix}$ to $\begin{pmatrix} 0 & AB \\ -I_n & 0 \end{pmatrix}$?

answered Jan 22 '23 at 13:35

Chia

1,023

can you please tell the transmission that you used ? – neelkanth Feb 10 '24 at 13:38
@neelkanth $$ \begin{pmatrix} I & A \ 0 & I \end{pmatrix} \begin{pmatrix} A & 0 \ -I & B \end{pmatrix} \begin{pmatrix} I & B \ 0 & I \end{pmatrix} = \begin{pmatrix} 0 & AB \ -I & 0 \end{pmatrix} $$ – Chia Feb 15 '24 at 03:01
Thank you ...... – neelkanth Feb 15 '24 at 03:19

Proving that $rank(A) + rank(B) - n \le rank(AB)$

1 Answers1