Proving Linear Independence in a System of Vectors

[annoymage]

Homework Statement

let u₁ , u₂ , u₃ $$\in$$ R³ be such that

Au₁^T = u₁^T , Au₂^T = $$\frac{1}{2}$$u₂^T , Au₃^T = $$\frac{1}{3}$$u₃^T

Suppose u₁ , u₂ , u₃ $$\neq$$ 0

Show that

u₁ , u₂ , u₃ are linearly independent

Homework Equations

a₁u₁ + a₂u₂ + a₃u₃ = 0

a₁ = a₂ = a₃ = 0 for it to become linearly independent

The Attempt at a Solution

Au_i^T = u_i^T iff u_i = u_i A^T for all i=1,2,3

implies

A^T( u₁ + $$\frac{1}{2}$$ u₂ + $$\frac{1}{3}$$ u₃) = 0

how to proof this is linear independent?
gimme clue please T_T

 

[Mark44]

Homework Statement

let u₁ , u₂ , u₃ $$\in$$ R³ be such that

Au₁^T = u₁^T , Au₂^T = u₂^T , Au₃^T = u₃^T

Suppose u₁ , u₂ , u₃ $$\neq$$ 0

Show that

u₁ , u₂ , u₃ are linearly independent

Homework Equations

a₁u₁ + a₂u₂ + a₃u₃ = 0

There's more to the definition of linear independence. For example, there is always a solution for a1, a2, a3 in the equation above no matter what vectors you have. It doesn't matter whether they're independent or dependent.

a₁ = a₂ = a₃ = 0 for it to become linearly independent

The Attempt at a Solution

Au_i^T = u_i^T iff u_i = u_i A^T for all i=1,2,3

implies

A^T( u₁ + $$\frac{1}{2}$$ u₂ + $$\frac{1}{3}$$ u₃) = 0

how to proof this is linear independent?
gimme clue please T_T

Here's a clue - what is the definition of linear independence?

 

[annoymage]

hmm, i still in "i don't know" owho..

there is always a solution for a1, a2, a3 in the equation above no matter what vectors you have. It doesn't matter whether they're independent or dependent.

im not good in english, so what you mean is that

a1 a2 a3 always have a solution because Av=0 , the v is always have solution, for all A and 0 inside matrix mxn. right?

and what i know the definition of linear independece is what i state

a₁u₁ + a₂u₂ + a₃u₃ = 0

a₁ = a₂ = a₃ = 0 for it to become linearly independent.

hehe, can give me other clue about other definition.. clueless right now, sorry T_T

 

[Mark44]

There is more to the definition of linear independence, and this is a subtlety that escapes very many students.

Here are two examples that illustrate what I'm talking about.

1. u = <1, 0, 0>, v = <0, 1, 0>, w = <1, 1, 0>
Consider the equation au + bv + cw = 0. Clearly a = b = c = 0 is a solution. Does that make these vectors linearly independent?

2. u = <1, 0, 0>, v = <0, 1, 0>, w = <1, 0, 1>
Consider the equation au + bv + cw = 0. Clearly a = b = c = 0 is a solution. Does that make these vectors linearly independent?

 

[Mark44]

hmm, i still in "i don't know" owho..



im not good in english, so what you mean is that

a1 a2 a3 always have a solution because Av=0 , the v is always have solution, for all A and 0 inside matrix mxn. right?

Your English is OK, and is probably better than my attempts to speak your language. We're not talking about matrices here; we're just talking about what it means for a set of vectors to be linearly dependent or linearly independent.

For vectors u, v, and w, and constants c1, c2, and c3, the equation c1u + c2v + c3w = 0 always has a solution for the constants c1, c2, and c3. No matter what the vectors u, v, and w are. It doesn't matter whether they're linearly dependent or linearly independent. That equation always has a solution. So there must be something else that allows us to make a distinction between linearly independent and linearly dependent vectors. That's what I'm trying to get you to realize.

and what i know the definition of linear independece is what i state

a₁u₁ + a₂u₂ + a₃u₃ = 0

a₁ = a₂ = a₃ = 0 for it to become linearly independent.

hehe, can give me other clue about other definition.. clueless right now, sorry T_T

 

[irycio]

Oh c'mon, the definition is very simple. Since I'm not familiar with typing the mathematic formulas yet, let me write it for 3 vectors, but it's easy to generalise:
x,y,z are linearly independent iff:
(ax+by+cz=0) =>( a=b=c=0)
In Mark44's example 1 there is another solution: a=b=1, c=(-1). However, there is no other solution for vectors shown in (2), hence they are independent.

Nothing subtle :P

E: Well, apparently Mark44 was faster in answering his own post :P.

 

[annoymage]

no, mark44's is correct. i realized something wait let me refrase my definition. its hard to use this symbols

 

[irycio]

I didtn't write he was incorrect. I just didn't agree with his calling the deffinition of linear independence "subtle".

 

[annoymage]

ooooooooooo, if that, hmm like at question 1st

1. (a,b,c) = (-t,-t,t) for all t (real number)

proving that this is linear dependent
but (0,0,0) is also a solution

hmm, i get it what you are trying to say..

let me refrase my definition

a₁u₁ + a₂u₂ + a₃u₃ = 0

for at least a_i $$\neq$$ 0 , i=1,2,3

then it is linearly dependent

otherwise, it is linear independent

am i right?

 

[annoymage]

owho, sorry irycio ;P

 

[annoymage]

but, how do i solve the question then? should i assume that the statement is linear independent and make contradiction?

 

[Mark44]

Oh c'mon, the definition is very simple. Since I'm not familiar with typing the mathematic formulas yet, let me write it for 3 vectors, but it's easy to generalise:
x,y,z are linearly independent iff:
(ax+by+cz=0) =>( a=b=c=0)

Nope, that's not it. Let's look at my first example.
u = <1, 0, 0>, v = <0, 1, 0>, w = <1, 1, 0>
I can see by inspection that au + bv + cz = 0 has a solution a = b = c = 0. Does that make vectors u, v, and w linearly independent? The answer is no.

In Mark44's example 1 there is another solution: a=b=1, c=(-1). However, there is no other solution for vectors shown in (2), hence they are independent.

Nothing subtle :P

I disagree. It's very subtle. Students always focus on the equation (e.g. for 3 vectors) au + bv + cz = 0 and completely miss the subtlety of the distinction between linear independence and linear dependence.

As I said before the equation always has a solution regardless of whether the vectors are linearly dependent or linearly independent. The subtlety comes with whether the solution for the constants is unique (a = b = c = 0) or not unique; i.e., that there are nonzero solutions for the constants. If the solution is unique, the vectors are linearly independent. If the solution is not unique, the vectors are linearly dependent.

E: Well, apparently Mark44 was faster in answering his own post :P.

 

[Mark44]

ooooooooooo, if that, hmm like at question 1st

1. (a,b,c) = (-t,-t,t) for all t (real number)

proving that this is linear dependent
but (0,0,0) is also a solution

This is a different question from the first one you posted.

(a, b, c) is a vector, and a single vector is linearly independent as long as it's not identically the zero vector. If u = <a, b, c> = <-t, -t, t> then the equation c1 u = 0 has only a single solution for arbitrary values of t, c1 = 0.

I'm not sure about what you're asking, though.

hmm, i get it what you are trying to say..

let me refrase my definition

a₁u₁ + a₂u₂ + a₃u₃ = 0

for at least a_i $$\neq$$ 0 , i=1,2,3

then it is linearly dependent

otherwise, it is linear independent

am i right?

 

[irycio]

Oh of course it always does have the zero-solution, but in my opinion it just is not definition.

What I wrote in thefragment which You first quoted means literally, that the vectors are independent ONLY IF zero is the ONLY solution. Your first example has another one (1,1,-1) and hence those vectors are NOT independent.

I believe we totally agree on the linear dependencies between vectors. It's just the definition I was told concentrated more on the uniqueness of the solution than on the equation SUM=0 itself (as important as it is, though .

 

[Mark44]

Oh of course it always does have the zero-solution, but in my opinion it just is not definition.

I don't understand what you're saying here.

What I wrote in thefragment which You first quoted means literally, that the vectors are independent ONLY IF zero is the ONLY solution.

That's not what you wrote, which was

(ax+by+cz=0) =>( a=b=c=0)

This is the part after iff. In my counterexample I supplied an equation ax + by + cz = 0 that is obviously true, and came up with a solution a=b=c=0 (which are true statements), so the implication is true. The problem is that according to your definition the vectors x, y, and z are linearly independent when in fact they are not.

Your first example has another one (1,1,-1) and hence those vectors are NOT independent.

Right. So your definition is defective. I recognize that you understand the difference between linear dependence/independence. All I'm saying is that your definition doesn't work. It should say that the equation ax + by + cz = 0 has exactly one solution for the constants a, b, and c. That is exactly the subtlety I've been talking about, and the one that has escaped many generations of linear algebra students.

I believe we totally agree on the linear dependencies between vectors. It's just the definition I was told concentrated more on the uniqueness of the solution than on the equation SUM=0 itself (as important as it is, though .

That is the correct definition - the uniqueness of the solution. The fact that a linear combination of vectors adds up to the zero vector is completely immaterial.

 

[irycio]

You're right, my definition "(ax+by+cz=0)..." didn't mean what I wanted it to mean.

 

[annoymage]

im still twisting my brain, how to do this,

hmm,

u₁(A^T - I) = 0

u¹$$\neq$$ 0 => (A^T - I) = 0

am I in the right path? ;P

 

[annoymage]

irycio, can you help me ? T_T

mark44, are you off9 ?

someone please , ngahaha I am begging you

 

[Mark44]

Do you know about eigenvalues, eigenvectors, and similar matrices? For this problem you have Au1 = u1, Au2 = (1/2)u2, and Au3 = (1/3)u3, where u1, u2, and u3 are not zero vectors.

You need to show that u1, u2, and u3 are linearly independent. I.e., show that the only solution of the equation c1u1 + c2u2 + c3u3 = 0 is c1 = c2 = c3 = 0.

To do this, you need to be able to say something about the matrix A.

 

[annoymage]

i see i see, let me do some reading, n then try solve it, after i go bath. owhoho..
my body all sticky, later if i get the proof, i'll post to let you check it.