Linear Independence of ci Given Linear Independence of ui?

[annoymage]

Homework Statement

let u_i , c_i \in Rⁿ

and

A \in M_n(R) be such that Ac_i^T = u_i^T , i=1,2,...,n

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

c₁,c₂,...,c_n are linearly independent

Homework Equations

N/A

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

 

[Mark44]

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.

Homework Statement

let u_i , v_i \in Rⁿ

and

A \in M_n(R) be such that Av_i^T = u_i^T , i=1,2,...,n

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

v₁,v₂,...,v_n are linearly independent

Homework Equations

N/A

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

Since u₁, u₂, ..., u_n are linearly independent, the equation c₁u₁ + c₂u₂ + ... + c_nu_n = 0 has only one solution for the constants c₁, c₂, ... , c_n. (What is that solution?)

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

 

[annoymage]

Since u₁, u₂, ..., u_n are linearly independent, the equation c₁u₁ + c₂u₂ + ... + c_nu_n = 0 has only one solution for the constants c₁, c₂, ... , c_n. (What is that solution?)

one solution which c_i= 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"

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

this one, since

Av_i^T = u_i^T

implies

u_i = v_iA^T

implies

c₁v₁A^T + c₂v₂A^T + ... + c_nv_nA^T = 0

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

implies

v₁A^T , v₂A^T ,..., v_nA^T are linear independent

am i right?? did i over complicating things?

if i right,

i only show v₁A^T , v₂A^T ,..., v_nA^T are independent

how to show

v₁ , v₂ ,..., v_n are independent

 

[annoymage]

can i make like this?

c₁v₁A^T + c₂v₂A^T + ... + c_nv_nA^T = 0

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

implies

(c₁v₁ + c₂v₂ + ... + c_nv_n)A^T = 0

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

implies

c₁v₁ + c₂v₂ + ... + c_nv_n = 0

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

implies

v₁ , v₂ ,..., v_n are independentim sorry if i do over complicating things and annoy you much. T_T

 

[Mark44]

one solution which c_i= 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"

Right. Just say that this solution is the only solution because it is given that the vectors are linearly independent.

this one, since

Av_i^T = u_i^T

implies

u_i = v_iA^T

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.

implies

c₁v₁A^T + c₂v₂A^T + ... + c_nv_nA^T = 0

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

implies

v₁A^T , v₂A^T ,..., v_nA^T are linear independent

am i right?? did i over complicating things?

Yes, a little, but you have the main idea. Instead of saying this:
c₁v₁A^T + c₂v₂A^T + ... + c_nv_nA^T = 0


you can instead say this:
c₁Av₁ + c₂Av₂ + ... + c_nAv_n = 0

All you have done is replace u_i in the first equation with Av_i 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.

if i right,

i only show v₁A^T , v₂A^T ,..., v_nA^T are independent

how to show

v₁ , v₂ ,..., v_n are independent

 

[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?

 

[Mark44]

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 v_i, Av_i = 0.

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

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