MHB Rank of the product of two matrices

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If A is an M by n matrix and B is an n by n matrix of rank n, then rank(AB) equals rank(A). This conclusion stems from the fact that an n by n matrix of rank n is invertible, allowing the use of the theorem that states rank(AB) is less than or equal to the minimum of the ranks of A and B. By applying the inverse of B, it can be shown that rank(A) is equal to rank((AB)B^{-1}), which leads to the conclusion that rank(AB) must also equal rank(A). The proof effectively demonstrates that the properties of matrix multiplication and rank are interconnected. Thus, the theorem is validated as a corollary to the established rank inequality.
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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