# Proving a Linear System

1. Apr 14, 2010

### annoymage

1. The problem statement, all variables and given/known data

let ui , ci $$\in$$ Rn

and

A $$\in$$ Mn(R) be such that AciT = uiT , i=1,2,...,n

Suppose u1,u2,....,un are linearly independent. Show that

c1,c2,....,cn are linearly independent

2. Relevant equations

N/A

3. The attempt at a solution

i was just learning the eigenvalue, eigenvector, but i didn't have any clue how to relate to this or this really related to eigen?

can someone help me

2. Apr 14, 2010

### Staff: Mentor

I hope you don't mind, but I changed the names of your c vectors to v. c is almost always used for constants, while u, v, w, etc are used for vectors.
Since u1, u2, ..., un are linearly independent, the equation c1u1 + c2u2 + ... + cnun = 0 has only one solution for the constants c1, c2, ... , cn. (What is that solution?)

Now, what can you say about the equation c1Av1 + c2Av2 + ... + cnAvn = 0?

3. Apr 14, 2010

### annoymage

one solution which ci= 0 , i = 1,2,....,n

right? i'm still not sure how to state that because, i remembered there are only one solution, but how to state that it is "only one solution"

this one, since

AviT = uiT

implies

ui = viAT

implies

c1v1AT + c2v2AT + ... + cnvnAT = 0

has only one solution ci = 0 , i=1,2,...,n

implies

v1AT , v2AT ,...., vnAT are linear independent

am i right?? did i over complicating things?

if i right,

i only show v1AT , v2AT ,...., vnAT are independent

how to show

v1 , v2 ,...., vn are independent

Last edited: Apr 14, 2010
4. Apr 14, 2010

### annoymage

can i make like this?

c1v1AT + c2v2AT + ... + cnvnAT = 0

has only one solution ci = 0 , i=1,2,...,n

implies

(c1v1 + c2v2 + ... + cnvn)AT = 0

has only one solution ci = 0 , i=1,2,...,n

implies

c1v1 + c2v2 + ... + cnvn = 0

has only one solution ci = 0 , i=1,2,...,n

implies

v1 , v2 ,...., vn are independent

im sorry if i do over complicating things and annoy you much. T_T

Last edited: Apr 14, 2010
5. Apr 15, 2010

### Staff: Mentor

Right. Just say that this solution is the only solution because it is given that the vectors are linearly independent.
I guess you took the transpose of each side. Not sure that this is worth doing. I would say that you can leave off the T's for transpose.
Yes, a little, but you have the main idea. Instead of saying this:
c1v1AT + c2v2AT + ... + cnvnAT = 0

c1Av1 + c2Av2 + ... + cnAvn = 0

All you have done is replace ui in the first equation with Avi in the second equation. Since the first equation has only one solution (the trivial solution c1=c2=...=cn=0), then so does the second equation.

6. Apr 15, 2010

### annoymage

hoho, thankyou very much mark44, hmm now only bugging me is the other question .. Let me try my best.

before that, can you tell whether A is invertible or not?

7. Apr 15, 2010

### Staff: Mentor

A is invertible.

If A were not invertible, then the dimension of the nullspace of A would have to be at least 1. That means that for at least one vector vi, Avi = 0.

Since you are given that Avi = ui, and the u vectors are linearly independent, none of them can be zero, hence Avi $\neq$ 0 for i = 1, 2, ..., n.

That's not the complete proof, but it should give you the idea.