Why are invertible matrices "one-to-one"?

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Answer 1 · 2015-12-20T01:11:49

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I think the question is referring to the natural use of a matrix to map points to points by multiplication.

Suppose $M$ is an invertible matrix with inverse $M^(-1)$

Suppose further that $Mp_1 = Mp_2$ for some points $p_1$ and $p_2$ .

Then multiplying both sides by $M^(-1)$ we find:

$p_1 = I p_1 = M^(-1)M p_1 = M^(-1)M p_2 = I p_2 = p_2$

So:

$Mp_1 = Mp_2 => p_1 = p_2$

That is: multiplication by $M$ is one-to-one.