Prove mutually non-zero orthogonal vectors are linearly independent

[unscientific]

Homework Statement

Let a₁, a₂, ... an be vectors in Rⁿ and assume that they are mutually perpendicular and none of them equals 0. Prove that they are linearly independent.

Homework Equations

The Attempt at a Solution

Consider β_ia_i + β_ja_j ≠ 0 for all i, j

=> β_ia_i + β_ja_j + β_ka_k ≠ 0 for all i, j, k.

Therefore β₁a₁ + β₂a₂ + ... + β_na_n ≠ 0 (Linearly independent)

 

[voko]

I do not understand your attempt.

 

[unscientific]

I do not understand your attempt.

I first start off with adding 2 vectors, and showing they are non-zero. Then I add a third one, which is also non-zero. Then i add everything to show it is also non-zero.

 

[voko]

I first start off with adding 2 vectors, and showing they are non-zero. Then I add a third one, which is also non-zero. Then i add everything to show it is also non-zero.

I do not see how you show that any of those sums is non-zero. You just state that it is. You might as well state the end result immediately, it will be just as (un)justified.

 

[Mark44]

Therefore β₁a₁ + β₂a₂ + ... + β_na_n ≠ 0 (Linearly independent)

This is NOT the definition of linear independence. The equation β₁a₁ + β₂a₂ + ... + β_na_n =[/color] 0 appears in the definition for linear independence, and in the definition for linear dependence.

How then do we distinguish between a set of vectors that is linearly independent from one that is linearly dependent?

What you showed above, with the ≠ symbol, doesn't appear in either definition.

 

[unscientific]

This is NOT the definition of linear independence. The equation β₁a₁ + β₂a₂ + ... + β_na_n =[/color] 0 appears in the definition for linear independence, and in the definition for linear dependence.

How then do we distinguish between a set of vectors that is linearly independent from one that is linearly dependent?

What you showed above, with the ≠ symbol, doesn't appear in either definition.

I think the first step is to show that the vector sum of any 2 vectors is non-zero. But since all the vectors are mutually orthogonal, sum of both can't be zero?

Quick proof:

Assume β_ia_i + β_ja_j = 0

This implies that a_i = -(β_j/β_i)a_j is parallel to a_j.
($\Rightarrow$$\Leftarrow$)
So any two mutually orthogonal vectors are linearly independent. By mathematical induction, Ʃβ_ia_i ≠ 0.

 

[hilbert2]

Consider the inner product of vector β₁a₁+β₂a₂+...+β_na_n with vector a_i and show that it is zero only if β_i=0.

Therefore β₁a₁+β₂a₂+...+β_na_n = 0 iff all β_i are zero.

 

[HallsofIvy]

I think the first step is to show that the vector sum of any 2 vectors is non-zero. But since all the vectors are mutually orthogonal, sum of both can't be zero?

Quick proof:

Assume β_ia_i + β_ja_j = 0

This implies that a_i = -(β_j/β_i)a_j is parallel to a_j.
($\Rightarrow$$\Leftarrow$)
So any two mutually orthogonal vectors are linearly independent. By mathematical induction, Ʃβ_ia_i ≠ 0.

You have not said anything about the $\beta_i$ not all being 0. Obviously if $\beta_i= 0$ for all i, that sum is 0.

 

[voko]

By mathematical induction, Ʃβ_ia_i ≠ 0.

You have, almost, proved the base of induction. But you still have to prove the $n \rightarrow n + 1$ induction step.

 

[unscientific]

You have, almost, proved the base of induction. But you still have to prove the $n \rightarrow n + 1$ induction step.

I don't think induction works here. Even though sum of any 2 vectors = 0, it doesn't mean that adding a third vector won't make it zero.

Assume the sum of 3 vectors = 0, it implies the third vector added is parallel to the sum of the two (No contradiction, as the statement says that the vectors are mutually orthogonal and nothing is said about the orthogonality between a vector and a sum of vectors.

I think the right way is to take the inner product of any vector with respect to the entire sum.

 

[voko]

I don't think induction works here. Even though sum of any 2 vectors = 0, it doesn't mean that adding a third vector won't make it zero.

Assume the sum of 3 vectors = 0, it implies the third vector added is parallel to the sum of the two (No contradiction, as the statement says that the vectors are mutually orthogonal and nothing is said about the orthogonality between a vector and a sum of vectors.

Well, this can in fact be proved, too, but this is probably more difficult than what you have to do.

I think the right way is to take the inner product of any vector with respect to the entire sum.

Why does that have to be any vector?

 

[unscientific]

Well, this can in fact be proved, too, but this is probably more difficult than what you have to do.



Why does that have to be any vector?

1. Add any 2 vectors, show that they are non-zero.

2. Add a third vector, take inner product of that sum with that vector just added. Non-zero.

3. Carry on process till last vector.

4. QED

 

[voko]

This does not prove linear independence. Use the definition of the latter.

 

[HallsofIvy]

Unscientific, the basic problem appears to be that you have an incorrect idea of what the definition of "independent" is!

It is NOT

1. Add any 2 vectors, show that they are non-zero.

2. Add a third vector, take inner product of that sum with that vector just added. Non-zero.

3. Carry on process till last vector.

4. QED

A set of vectors $$\{v_1, v_2, ..., v_n\}$$ is "independent" if an only if the only way $$\beta_1v_1+ \beta_2v_2+ ... + \beta_nv_n= 0$$ is if $$\beta_1= \beta_2= ...= \beta_n$$.

It is NOT just a matter of adding vectors and saying the sum is not 0. It is the the only way a linear combination of them can be 0 is if all coefficients are 0. That is equivalent to the statement that no one of the vectors can be written as a linear combination of the other.

So to prove a set of vectors is linearly independent is to start, "Suppose $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n= 0$$" and show that every one of the \betas is equal to 0. Here the only condition on the vectors is that they are "mutually perpendicular"- and no where have you used that condition.

What would you get if you took the dot product of $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n$$ with each of $$v_1$$, $$v_2$$, ..., $$v_n$$ in turn?

 

[unscientific]

Unscientific, the basic problem appears to be that you have an incorrect idea of what the definition of "independent" is!

It is NOT


A set of vectors $$\{v_1, v_2, ..., v_n\}$$ is "independent" if an only if the only way $$\beta_1v_1+ \beta_2v_2+ ... + \beta_nv_n= 0$$ is if $$\beta_1= \beta_2= ...= \beta_n$$.

It is NOT just a matter of adding vectors and saying the sum is not 0. It is the the only way a linear combination of them can be 0 is if all coefficients are 0. That is equivalent to the statement that no one of the vectors can be written as a linear combination of the other.

So to prove a set of vectors is linearly independent is to start, "Suppose $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n= 0$$" and show that every one of the \betas is equal to 0. Here the only condition on the vectors is that they are "mutually perpendicular"- and no where have you used that condition.

What would you get if you took the dot product of $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n$$ with each of $$v_1$$, $$v_2$$, ..., $$v_n$$ in turn?

Yup, sorry for not being concise. What i meant by "add any 2 vectors" I mean adding β_ia_i.

 

[unscientific]

Unscientific, the basic problem appears to be that you have an incorrect idea of what the definition of "independent" is!

It is NOT A set of vectors $$\{v_1, v_2, ..., v_n\}$$ is "independent" if an only if the only way $$\beta_1v_1+ \beta_2v_2+ ... + \beta_nv_n= 0$$ is if $$\beta_1= \beta_2= ...= \beta_n$$.

It is NOT just a matter of adding vectors and saying the sum is not 0. It is the the only way a linear combination of them can be 0 is if all coefficients are 0. That is equivalent to the statement that no one of the vectors can be written as a linear combination of the other.

So to prove a set of vectors is linearly independent is to start, "Suppose $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n= 0$$" and show that every one of the \betas is equal to 0. Here the only condition on the vectors is that they are "mutually perpendicular"- and no where have you used that condition.

What would you get if you took the dot product of $$\beta_1v_1+ \beta_2v_2+ ...+ \beta_nv_n$$ with each of $$v_1$$, $$v_2$$, ..., $$v_n$$ in turn?

1. Add any 2 vectors, show that they are non-zero.

β_ia_i + β_ja_j can't be zero, otherwise

a_i = -(β_j/β_i)a_j implying they are parallel. Contradiction, as they are orthogonal.2. Add a third vector, take inner product of that sum with that vector just added. Non-zero.

β_ia_i + β_ja_j + β_ka_k.

Taking inner product, the first two innerproducts give 0, due to orthogonality. The last one, which is with itself, gives non-zero due to positivity of the norm.

3. Carry on process till last vector.

4. QED (assuming coefficients are non-zero)

I hope this is clear enough..thanks for the help guys!

 

[micromass]

1. Add any 2 vectors, show that they are non-zero.

β_ia_i + β_ja_j can't be zero, otherwise

What if $\beta_i = \beta_j = 0$. Doesn't the expression give $0$?

a_i = -(β_j/β_i)a_j implying they are parallel. Contradiction, as they are orthogonal.

What if $\beta_i=0$, won't you divide by $0$?
Why are parallel vectors not orthogonal?

 

[unscientific]

What if $\beta_i = \beta_j = 0$. Doesn't the expression give $0$?



What if $\beta_i=0$, won't you divide by $0$?
Why are parallel vectors not orthogonal?

I'm assuming all coefficients are non-zero.

 

[micromass]

I'm assuming all coefficients are non-zero.

Well, you need to say this. And why can you assume this anyway?

 

[unscientific]

Well, you need to say this. And why can you assume this anyway?

Because I am choosing them to be non-zero, in order to work towards the proof of linear independence. No point choosing any of them to be zero. (I thought this was straightforward enough not to say..)

 

[micromass]

Because I am choosing them to be non-zero, in order to work towards the proof of linear independence. No point choosing any of them to be zero. (I thought this was straightforward enough not to say..)

That's not what linear independence states. It says that $\beta_i\mathbf{v}_i + \beta_j\mathbf{w}_j \neq \mathbf{0}$ whenever $\beta_i$ and $\beta_j$ are both nonzero. So it can certainly happen that one of the $\beta_i$ is zero.

 

[unscientific]

That's not what linear independence states. It says that $\beta_i\mathbf{v}_i + \beta_j\mathbf{w}_j \neq \mathbf{0}$ whenever $\beta_i$ and $\beta_j$ are both nonzero. So it can certainly happen that one of the $\beta_i$ is zero.

Yes I get what you mean. But the question wants to show the linear independence of all vectors from a₁ to a_n! If you let any of the coefficients be 0, then you are at most showing linear independence of all vectors except that one you let the coefficient be 0.

 

[voko]

Yes I get what you mean. But the question wants to show the linear independence of all vectors from a₁ to a_n! If you let any of the coefficients be 0, then you are at most showing linear independence of all vectors except that one you let the coefficient be 0.

You are required to prove that the entire set of vectors is linearly independent. By definition, you must prove that their linear combination is zero only when all the coefficients are zero. If you prove any other statement, then you need also a proof that your statement leads to linear independence per the original definition.

 

[unscientific]

You are required to prove that the entire set of vectors is linearly independent. By definition, you must prove that their linear combination is zero only when all the coefficients are zero. If you prove any other statement, then you need also a proof that your statement leads to linear independence per the original definition.

That's right, thanks for putting it in a more elegant way!