Left multiplication by invertible matrix. If M is an $n\times m$ matrix of rank $n$ and $A\in GL_n$, that is, $A$ is an invertible $n\times n$ matrix, then why is the row reduced echelon form of $M$ is the same as the row reduced echelon form of $AM$?
 A: I will not answer your question directly, but rather give you hints by referring to the relevant definitions and theorems:


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*First, be sure to understand that the row reduced echelon form of a matrix is unique (Refer for example to Matrices and linear transformations by Cullen, theorem 2.18), but a matrix in row reduced echelon form might be row equivalent to more than one matrix (equivalence relation - so several matrices may belong to the same row equivalent class):— so $M$ and $AM$ will generally not be the same matrix, but they could be row equivalent to the same matrix in row reduced echelon form.  

*The matrix $M$ is row equivalent to a matrix in row reduced echelon form (say this matrix is $N$) iff there exists an invertible matrix $P$ such that $PM=N$. Also, every $n \times m$ matrix over a field is row equivalent to a matrix in row reduced echelon form (so we are guaranteed that such a $P$ and $N$ exist). For the theory related to these statements please refer to, for example, section 1.8 and 1.9 in the same text as mentioned above (this is all about row equivalence, series of elementary operations, elementary matrices, etc. it helps to know this).

*Now take $AM$: if $PM=N$ with $N$ in row reduced echelon form, can you find an invertible matrix $Q$ such that $QAM=N$? By finding $Q$ you will prove that $M$ and $AM$ have the same row reduced echelon form.


I hope that is enough guidance? Ask if you want clarification on something specific.
