What is the relationship between invertible linear mappings and rank in proofs?

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The discussion focuses on the relationship between invertible linear mappings and rank in linear algebra. It establishes that if M is an invertible linear mapping from ℝm to ℝm, then the rank of the composition (M o L) is equal to the rank of L. This is due to the fact that an invertible mapping preserves the dimensionality of the image space, ensuring that the rank remains unchanged. The dimension of the null space of M is zero, confirming that the rank of M is m.

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Mona1990
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1. Hi!
I was wondering if anyone could help me to solve the following problem!
Let L : [R][n] ->[R][m] and M :[R][m]-> [R][m] be linear mappings.
Prove that if M is invertible, then rank (M o L) = rank (L)

thanks! :)
 
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If M is invertible it maps Rm one-to-one onto Rm. In particular, it maps any k dimensional subset of Rm onto a k dimensional subset of Rm. Now, what does "rank" mean?
 
the dimension of the column space of M is the rank of M

and we know that dim (null M)= 0 since the null space of M is just the zero vector
and since rank = m - (dimension of the null space of M)
so rank is m?
 

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