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Rank of product of matrices

Linear Algebra
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Riemann
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Rank of product of matrices

#1

Post by Riemann » Sun Jun 11, 2017 9:17 pm

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

\[{\rm rank} (AB) \geq {\rm rank} (A) + {\rm rank}(B) -n\]
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Tolaso J Kos
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Re: Rank of product of matrices

#2

Post by Tolaso J Kos » Fri Nov 06, 2020 11:59 am

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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