## Rank of product of matrices

### 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

\[{\rm rank} (AB) \geq {\rm rank} (A) + {\rm rank}(B) -n\]

\[{\rm rank} (AB) \geq {\rm rank} (A) + {\rm rank}(B) -n\]

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

- Tolaso J Kos
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**Posts:**866**Joined:**Sat Nov 07, 2015 6:12 pm**Location:**Larisa-
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### 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.**

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