# Proofs for Invertible Matrix Theorems

a->k

If A is an invertable matrix then there exists a matrix $A^{-1}$ for which $A\cdot A^{-1}=I$

Therefore there exists some matrix D (ie the inverse) for which $A\cdot D=I$

k->j

If there exists some matrix D for which $A\cdot D=I$

D must be the inverse.

If D is the inverse then there must exist some matrix C for which $C\cdot A=I$ because $A\cdot A^{-1}=I=A^{-1} \ cdot A$

c>d

If there are n pivot rows in an nxn matrix then when you augment it with the 0 vector there is no way for the solution to equal anything but the 0 vector.

d>e

If there exists only the trivial solution for Ax=0 then the columns can be reduced to the Identity matrix. If it can be reduced to the Identity matrix then the columns must form a linearly independent set because they are not multiples of each other and there is no 0 column.