Hi all; I have two quick Linear Algebra questions that I am struggling with. I think they are connected.
#1) Suppose CA = I(sub)n (the n * n identity matrix). Show that the equation Ax = 0 has only the trivial solution. Explain why A cannot have more columns than rows.
My thoughts: For the second half, I believe this is because if it had more columns than rows, there would be free variables, which would mean that the equation was dependent and not only the trivial solution exists. I can't figure out the proof, however.
#2) Suppose A is an m * n matrix and there exist n * m matrices C & D such that CA = I(sub)n and AD = I(sub)m. Prove that m = n and C = D. [Hint: Think about the product CAD].
My thoughts: I don't erally understand the whole In and Im business, for starters. This question has me lost.
Thanks for the help!!!
Feb 17th 2010, 06:44 PM
CA = I
==> A is invertible
==> Ax = 0 has only the trivial solution
since Ax = 0
=> A^-1 A x = A^-1 (0) = x
==> 0 = x
if A wasn't invertible then you couldn't make that distinction in general
CAD = I_n D = D
D^-1 CAD = I_m
D^-1 CAD = D^-1 C (AD) = D^-1 C I_m
==> D^-1C = I_m
==> C = D
and you're done