Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Rank of product of matrices

Linear Algebra
Riemann
Articles: 0
Posts: 170
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

### Rank of product of matrices

Let $A, B$ be $m \times n$ and $n \times k$ matrices respectively with entries over some field. Prove that

${\rm rank} (AB) \geq {\rm rank} (A) + {\rm rank}(B) -n$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tolaso J Kos
Articles: 2
Posts: 860
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

### Re: Rank of product of matrices

Lemma wrote:It holds that

$${\rm nul} (T_1 T_2) \leq {\rm nul} (T_1) + {\rm nul} (T_2)$$

where $T_1, \; T_2$ are the corresponding linear transformations.
Proof: The proof of the lemma is based on the rank - nullity theorem.

Based upon the above lemma we have that

\begin{align*}
{\rm rank} \left ( T_1 T_2 \right ) + n &= k - {\rm nul} \left ( T_1 T_2 \right ) +n \\
&\geq n - {\rm nul} (T_1) + k - {\rm nul} (T_2) \\
&={\rm rank} (T_1) + {\rm rank} (T_2)
\end{align*}
Imagination is much more important than knowledge.