Answer 1

I was able to prove the direction "If A is invertible, then {Av1, ... , Avn} is a basis for R^n". But I am having a lot of trouble proving the other direction (If {Av1, ..., Avn} is a basis, then A is invertible.

I was thinking the easiest way to do it would be to show that rank A = n, and then A would be invertible. However, I am having trouble showing that rank A = n...

Answer 2

if A is invertible then its row equivalent to the identity matix

Answer 3

if we know that the set is a basis for A, then we know it is linearly independant, and therefore has a nonzero determinant

Answer 4

and since Avn has n columns; and exists in R^n it has n rows making an nxn square

Answer 5

Oh really if vectors are linearly independent, then det A is not zero? I'm not sure if we learnt that o.O

Answer 6

if yor on basis, then yes, youve learnt that lol

Answer 7

det=0=dep; det not0 = not dep

Answer 8

Hm ok looking more through my book, there's only an end-of-chapter problem saying if they;re lin. independent, then det = not zero. But we didn't see it formally beforehand. I'm not sure if I can use it as a theorem

Answer 9

then use it the long way :)

Since ..Avn is a basis then it only has the trivial solution Ax = 0

Answer 10

or rather; Avn is invertible if it only has the trivial solution for Ax = 0

Answer 11

something along those lines ... all those thrms and such mean th esame thing in the end

Answer 12

Oh I see. Ok i think that will work better

Answer 13

i hope so, cause thats all i can really remember of that stuff at the moment lol