rang inequality 2

hxtung · Riemann Hypothesis Offline Joined: 26 Jul 2003 Posts: 323

Let $P,Q,R \in M_n(K)$ . Prove that $rang(PQ)+rang(QR) \leq rang(Q)+rang(PQR)$

**Posted:** Sat Apr 16, 2005 7:16 pm

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liyi

what is 'rang'?
Range or Rank?

**Posted:** Sat Apr 16, 2005 7:39 pm

hxtung · Riemann Hypothesis Offline Joined: 26 Jul 2003 Posts: 323

rang it means rank

**Posted:** Sat Apr 16, 2005 7:52 pm

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The heart of Mathematics is its problems

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

It is Frobenius inequality:
Firstly let's prove useful inequality,considering our matrices as mapping on vector space over $K$ .Let $U \subset V$ and $X$ : $V \to W$ . Then $dim(KerX|_{U})\leq dimKerX=dimV-dimImX$ (1). If we use inequality (1) for the case $U=ImQR$ , $V=ImQ$ , $X=P$ we obtain that $dim(KerP|_{ImQR})\leq dimImQ-dimImPQ$ (2). On the other hand it is readily that $dim(KerP|_{ImQR})=dimImQR-dimImPQR$ (3). So, applying (2) and (3) we get our inequality

**Posted:** Sat Apr 16, 2005 8:19 pm

liyi

A different approach.

$\begin{pmatrix} PQ & 0\\ Q & QR \end{pmatrix}$
Multiply the first column by $-R$ to the right and add it to the second column, then multiply the second column by $-1$ ,
$\begin{pmatrix} PQ & PQR\\ Q & 0 \end{pmatrix}$
Multiply the second row by $-A$ to the left, and add it to the first row
$\begin{pmatrix} 0 & PQR\\ Q & 0 \end{pmatrix}$

$rank(PQ)+rank(QR)=rank \begin{pmatrix} PQ & 0\\ 0 & QR \end{pmatrix} \leq rank \begin{pmatrix} PQ & 0\\ Q & Q...$

**Posted:** Sat Apr 16, 2005 9:14 pm