Rank of the product of two matrices

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SUMMARY

The discussion focuses on proving that if A is an M by n matrix and B is an n by n matrix of rank n, then rank(AB) = rank(A). This statement is established as a corollary to the theorem stating that for any two matrices A and B that can be multiplied, rank(AB) is less than or equal to the minimum of rank(A) and rank(B). The proof leverages the fact that an n by n matrix of rank n is invertible, allowing the use of the inverse B-1 to demonstrate that rank(A) equals rank(AB).

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aukie
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Hello

Both of the below theorems are listed as properties 6 and 7 on the wikipedia page for the rank of a matrix.

I want to prove the following,

If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A).

Apparently this is a corollary to the theorem
If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ).

which I know how to prove. But I can't prove the first theorem. Any ideas? I would especially like to see how it is a corollary to the second theorem which the author in the book I am reading claims. Thanks for reading
 
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aukie said:
I want to prove the following,

If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A).

Apparently this is a corollary to the theorem
If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ).
You want to prove that if A is an M by n matrix and B is an n by n matrix of rank n, then rank(AB) = rank(A). But an n by n matrix of rank n is necessarily invertible. So $B$ has an inverse $B^{-1}$. It follows from the theorem that $\text{rank}(A) = \text{rank}((AB)B^{-1}) \leqslant \text{rank}(AB).$ The reverse inequality $\text{rank}(AB)\leqslant \text{rank}(A)$ follows directly from the theorem. Hence $\text{rank}(AB)= \text{rank}(A).$
 

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