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Summary
Let A be an n x n matrix. A is invertible if and only if
The columns of A span R
n
For every b in R
n
, the system Ax=b is consistent
The rank of A is n
The columns of A are linear independent
The only solution to Ax=0 is the zero vector
The nullity of A is zero
The reduced row echelon form of A is I
n
A is a product of elementary matrices
There exists an n x n matrix B such that BA = I
n
There exists an n x n matrix C such that AC = I
n
onto
One-to-one
=
square matrix
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