Linear Independence of ci Given Linear Independence of ui?

In summary, a linear system is a set of linear equations that are solved simultaneously to find the values of the variables that satisfy all the equations. There are three methods for proving a linear system: substitution, elimination, and graphing. The solution to a linear system is the set of values for the variables that make all the equations in the system true. A linear system can have one, infinite, or no solutions, depending on the relationship between the equations. Proving a linear system is important because it allows us to find solutions that have real-life applications and helps us understand the relationships between variables.
  • #1
annoymage
362
0

Homework Statement



let ui , ci [tex]\in[/tex] Rn

and

A [tex]\in[/tex] 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

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
 
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  • #2
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.
annoymage said:

Homework Statement



let ui , vi [tex]\in[/tex] Rn

and

A [tex]\in[/tex] Mn(R) be such that AviT = uiT , i=1,2,...,n

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

v1,v2,...,vn 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, ..., 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
Mark44 said:
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?)

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"

Mark44 said:
Now, what can you say about the equation c1Av1 + c2Av2 + ... + cnAvn = 0?

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
 
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  • #4
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 independentim sorry if i do over complicating things and annoy you much. T_T
 
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  • #5
annoymage said:
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"
Right. Just say that this solution is the only solution because it is given that the vectors are linearly independent.
annoymage said:
this one, since

AviT = uiT

implies

ui = viAT
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.
annoymage said:
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?
Yes, a little, but you have the main idea. Instead of saying this:
c1v1AT + c2v2AT + ... + cnvnAT = 0


you can instead say this:
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.

annoymage said:
if i right,

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

how to show

v1 , v2 ,..., vn are independent
 
  • #6
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
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 [itex]\neq[/itex] 0 for i = 1, 2, ..., n.

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

FAQ: Linear Independence of ci Given Linear Independence of ui?

What is a linear system?

A linear system is a set of linear equations that are solved simultaneously to find the values of the variables that satisfy all the equations. Each equation in the system contains variables that are raised to the first power and are not multiplied together.

How do you prove a linear system?

To prove a linear system, you can use one of three methods: substitution, elimination, or graphing. Substitution involves solving one equation for one variable and substituting that expression into the other equations. Elimination involves adding or subtracting equations to eliminate one variable and solve for the remaining variables. Graphing involves graphing each equation and finding the point of intersection, which is the solution to the system.

What is the solution to a linear system?

The solution to a linear system is the set of values for the variables that make all the equations in the system true. This is also known as the point of intersection for the equations when graphed on a coordinate plane.

Can a linear system have more than one solution?

Yes, a linear system can have one, infinite, or no solutions. If the equations in the system are parallel, there is no point of intersection and no solution. If the equations are the same, there are infinite solutions because all points on the line are solutions. If the equations intersect at one point, there is one unique solution.

Why is proving a linear system important?

Proving a linear system is important because it allows us to find the solution to a system of equations, which can have real-life applications in fields such as engineering, physics, and economics. It also helps us understand the relationships between variables and how they affect each other.

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