nxn Matrix and Proof
Date: Winter 2013-2014
I was given a bunch of equivalent statements when given that an n x n matrix A is invertible. One of the statements said that if the first statement is true, then A is row equivalent to the nxn identity matrix. I have looked at a few proofs, but they are a bit confusing. Is there a proof by contradiction or an easier way to explain it?
Thanks for the question. First, I would like to suggest taking a look at Anton's Linear Algebra book. It is a classic in elementary linear algebra. This book has all the equivalent statements for nxn invertible matrices as well as their proofs. It is an excellent resource to have. Basically, here is the sketch of the proof: If A is nxn and invertible, it can be converted to the nxn identity matrix by elementrary row operations. Elementary row operations are what is meant by "row equivalence". In order to invert an nxn matrix, you must follow the (very tedious) algorithm of using elementary row operations on the given matrix A to get it to the nxn identity matrix. Thus, if a nxn matrix A is invertible, it must be row equivalent to the nxn identity matrix. Personally, I use the fact that det(A) not equaling zero as the test to show that a matrix is invertible.
I hope this helps.
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