# Orthogonal Basis and Inner Products

• TranscendArcu
In summary: At this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be.
TranscendArcu

## The Attempt at a Solution

Since A is a vector in V and since the $A_i$ form a basis, we can write A as a linear combination of the $A_i$. We write $A = x_1 A_1 + ... + x_n A_n$. Thus, we have,

$<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>$. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

$0 = x_i <A_i,A_i>$. Because we have presumed that the $A_i ≠ 0$, we cannot have $<A_i,A_i> = 0$, so we must have that the $x_i = 0$. So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice? TranscendArcu said: ## Homework Statement ## The Attempt at a Solution Since A is a vector in V and since the [itex]A_i$ form a basis, we can write A as a linear combination of the $A_i$. We write $A = x_1 A_1 + ... + x_n A_n$. Thus, we have,

$<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>$. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

$0 = x_i <A_i,A_i>$. Because we have presumed that the $A_i ≠ 0$, we cannot have $<A_i,A_i> = 0$, so we must have that the $x_i = 0$.

you're fine up to here.

So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0$. Which shows that A = 0.

I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice?

at this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be.

TranscendArcu said:

## The Attempt at a Solution

Since A is a vector in V and since the $A_i$ form a basis, we can write A as a linear combination of the $A_i$. We write $A = x_1 A_1 + ... + x_n A_n$. Thus, we have,

$<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>$. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

$0 = x_i <A_i,A_i>$. Because we have presumed that the $A_i ≠ 0$, we cannot have $<A_i,A_i> = 0$,
You can say more than that- since you are told that the basis is "orthonormal", you know that $<A_i, A_i>= 1$

so we must have that the $x_i = 0$. So we have $<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0. I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice? Once you have [itex]x_i= 0$ for all x, you have immediately that $A= 0A_1+ 0A_2+ ...= 0$

Okay. That makes sense. Thank you both.

This problem is related to the idea of the inner product. I don't think I need any help on the first parts, but I'd like to talk about the third part of the problem.

So, finding the length of a given vector given this inner product:
$<(x,y),(x,y)> = 5x^2 + y^2$.

Taking the length, we have

$|(x,y)| = \sqrt{5x^2 + y^2}$, which we define as equaling 1. Squaring both sides we find,

$5x^2 + y^2 = 1$. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at .5,-.5.

Is this correct?

## 1. What is an orthogonal basis?

An orthogonal basis is a set of vectors in a vector space that are all mutually perpendicular to each other, meaning they form right angles. In other words, the dot product of any two vectors in an orthogonal basis is equal to 0. This type of basis is important in linear algebra because it allows for easier calculations and understanding of vector operations.

## 2. How is an orthogonal basis related to inner products?

An inner product is a mathematical operation that takes in two vectors and produces a scalar value. In the context of an orthogonal basis, the inner product is used to determine the coefficients of a vector when it is expressed in terms of the basis vectors. This is because the inner product of a vector with each basis vector is equal to the coefficient of that basis vector in the vector's representation.

## 3. What are some examples of orthogonal bases?

Some examples of orthogonal bases include the standard basis for Euclidean space, where the basis vectors are the unit vectors along the x, y, and z axes, and the Fourier basis, which is used in signal processing and has orthogonal sine and cosine functions as basis vectors. Other examples include the Legendre polynomials and the Chebyshev polynomials, which are used in approximation and interpolation problems.

## 4. How is an orthogonal basis used in data analysis?

In data analysis, an orthogonal basis can be used to simplify and reduce the dimensionality of a dataset. This is achieved by projecting the data onto the basis vectors, which allows for easier visualization and interpretation of the data. Additionally, orthogonal bases are used in techniques such as principal component analysis (PCA) and singular value decomposition (SVD) to identify important features and patterns in the data.

## 5. What is the relationship between orthogonal bases and linear independence?

An orthogonal basis is always a linearly independent set of vectors. This means that none of the basis vectors can be expressed as a linear combination of the other basis vectors. This is because, in an orthogonal basis, each vector is perpendicular to all the others, and therefore cannot be a multiple of any other vector in the basis. The concept of linear independence is important in linear algebra because it allows for unique solutions to linear systems of equations.

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