Can someone give me a proof for the following statement?
$B$ is an invertible $n\times n$ matrix, then the rank of $AB$ is the same as the rank of $A$ for every $m\times n$ matrix $A$. Is the converse true?
Thank you all
Asked
Active
Viewed 974 times
1 Answers
0
Because $B$ is invertible then $B$ is nonsingular so $rank(B)=n$. Now we know that the rank of $CD$ is less than or equal to the minimun between $rank(C)$ and $rank(C)$ (http://en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties). Can you continue from here?
The converse is not true. What happen if $rank(A)<rank(B)<n$.
ILikeMath
- 1,514
-
How do you prove that rank of $CD$ is equal to the minimum between $rank(C)$ and $rank(D)$? – Mark Sep 28 '13 at 03:21
-
It is a standard result: http://en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties. – ILikeMath Sep 28 '13 at 03:22
-
Alright, then converse may not be true if we set B as $n \times n$ singular matrix which has rank that is higher than A but less than $n$? – Mark Sep 28 '13 at 03:27
-
-