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I have to show following things and have no idea what I have to do at b)

a) $GL(n,K) \times K^{n \times m} \rightarrow K^{n \times m}: (g,A) \mapsto gA$ is a group action.
b) The Rank is an invariant of this group action.
c) Rank is a separating invariant (Proof/Counter Example)

a) With the identity matrix $E_n \in GL(n,K)$ follows: $E_n \cdot A = A$ for all $A \in K^{n \times m}$. And since matrix multiplication is associative, $(gh)m = g(hm)$ for all $g,h \in GL(n,K)$ and all $m \in K^{n \times m}$.

How do I show, that the rank is an invariant?

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How to solve c)? – David Chan Mar 26 at 12:59

1 Answer 1

up vote 1 down vote accepted

Part b) is equivalent to the following: Given an $n \times m$ matrix $B$ and an invertible $n \times n$ matrix $A$, prove that the rank of $AB$ is equal to the rank of $B$. Do you know this fact?

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Ok, that helps profoundly. – monoid Oct 30 '12 at 14:34

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