# Prove Finite Orthogonal Set is Linearly Independent

• bugatti79
In summary, the conversation discusses proving that a finite orthogonal set is linearly independent. The proof involves using the fact that the set is orthogonal and showing that the only term in the sum that is not necessarily 0 is a_j <x_j, x_j>, which is equal to 1 if the set is orthonormal.
bugatti79
Folks,

I am looking at my notes. Wondering where the highlighted comes from.
Prove that a finite orthogonal set is lineaarly independent

let u=(x_1,x_2,x_n) bee an orthogonal set set of vectors in an ips.
To show u is linearly independent suppose

Ʃ ##\alpha_i x_i=0## for i=1 to n

Fix any j=1 and consider <Ʃ##\alpha_i x_i, x_j##> i=1 to n

then

0=<Ʃ##\alpha_i x_i, x_j##> i=1 to n

=Ʃ<##\alpha_i x_i, x_j##> i=1 to n

=Ʃ##\alpha_i <x_i, x_j>## i=1 to n

=##\alpha_j <x_j, x_j>## since u is an orthonormal set

Where does this line come from? Thanks

You are assuming the x_i's are orthogonal (i.e. <x_i,x_j>=0 if i =/= j), so the only term in the sum which is not necessarily 0 is a_j<x_j,x_j>

TwilightTulip said:
You are assuming the x_i's are orthogonal (i.e. <x_i,x_j>=0 if i =/= j), so the only term in the sum which is not necessarily 0 is a_j<x_j,x_j>

where <x_j,x_j>=1 when i=j? Thanks in advance.

Not necessarily. If you were to use "orthonormal" instead of just "orthogonal", then that would be true.

However, that is not necessary to your proof. $<x_j, x_j>$ is some non-zero number. $a_j$. Divide both sides of $\alpha_j a_j= 0$ by that number to get $\alpha_j= 0$.

.

The line comes from the fact that since u is an orthogonal set, the inner product of any two distinct vectors in the set is equal to 0. Therefore, when we take the inner product of the sum of all the vectors and any individual vector in the set, all terms except for the one with the same vector will be equal to 0. This leaves us with only the inner product of that same vector with itself, which is equal to 1 since u is an orthonormal set. Therefore, we end up with only the coefficient of that vector, which is equal to 0 since the sum of all the vectors is equal to 0. This shows that all the coefficients in the linear combination are equal to 0, proving that the set is linearly independent.

## 1. What is a finite orthogonal set?

A finite orthogonal set is a collection of vectors in a vector space that are all mutually perpendicular to each other. This means that the dot product of any two vectors in the set is equal to 0.

## 2. How do you prove that a finite orthogonal set is linearly independent?

To prove that a finite orthogonal set is linearly independent, we must show that any linear combination of the vectors in the set that equals 0 must have all coefficients equal to 0. This can be done by using the definition of a finite orthogonal set and setting up a system of equations to solve for the coefficients.

## 3. Can a finite orthogonal set be linearly dependent?

No, a finite orthogonal set cannot be linearly dependent. This is because the definition of a finite orthogonal set states that all vectors in the set must be mutually perpendicular, meaning that there is no way to create a linear combination that equals 0 without all coefficients being equal to 0.

## 4. Are there any real-world applications of proving a finite orthogonal set is linearly independent?

Yes, there are many real-world applications of this concept. One example is in signal processing, where orthogonal functions are used to represent signals. By proving that the set of orthogonal functions is linearly independent, we can ensure that the signal can be accurately reconstructed using these functions.

## 5. Can a finite orthogonal set be both linearly independent and linearly dependent?

No, a finite orthogonal set cannot be both linearly independent and linearly dependent. These two concepts are mutually exclusive, meaning that a set cannot possess both properties simultaneously.

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