Rank of product of matrices

Linear Algebra
Post Reply
User avatar
Posts: 178
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Rank of product of matrices


Post by Riemann »

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

\[{\rm rank} (AB) \geq {\rm rank} (A) + {\rm rank}(B) -n\]
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
User avatar
Tolaso J Kos
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa

Re: Rank of product of matrices


Post by Tolaso J Kos »

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

{\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)
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute


Sign in

Who is online

Users browsing this forum: No registered users and 0 guests