# Inner product space and orthonormal basis.

• Locoism
In summary, you have found the orthonormal basis for W and Wperp. You have found that Wperp has 2 vectors.

## Homework Statement

Assume the inner product is the standard inner product over the complexes.
Let W=
Spanhttp://img151.imageshack.us/img151/6804/screenshot20111122at332.png [Broken]

Find an orthonormal basis for each of W and Wperp..

## The Attempt at a Solution

Obviously I need to use Gram-Schmidt orthogonalization here, to find the orthonormal basis.
So I've applied the process to W, and normalized each vector to get an orthonormal basis. But I'm confused here as to what I would do for Wperp.. Don't I use G-S on W to find Wperp.? Am I finding the basis of W or am I finding Wperp. by applying G-S, and how am I supposed to find the other one?
If my guess is correct, do I find Wperp. by applying G-S to W, and then the orthonormal basis is each of these vectors normalized? Then I would apply G-S to Wperp. and normalized each of those vectors.

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Gram schimdt uses a set of linearly independent vectors to generate a set of orthornormal vectors with teh same span.

You have 3 vectors that span W. Are they linearly independent? If so use gram-schmidt to generate a set of 3 orthonormal vectors that span W.

Now consider W perp. As W is spanned by 3 linearly independent vectors (check?), Wperp will be spanned by 2 linearly independent vectors. Note every vector in W perp is orthogonal to the all vectors in W. You could use this fact with dot products to find basis vectors for Wperp.

How am I supposed to check they are linearly independent? Besides that, if they aren't, how would it be possible to use gram-schmidt? Also, how come Wperp will only have 2 vectors? I think I'm missing what you mean when you say use the dot product. There are 6 possible combinations if Wperp has 2 vectors, and each one would equal zero. But wouldn't it take ridiculously long to write them all down and work out each one?

Locoism said:
How am I supposed to check they are linearly independent?
you should probably understand how to check for linear independence prior to using gram-schmidt
Locoism said:
Besides that, if they aren't, how would it be possible to use gram-schmidt?
Actually you can still attempt gram schimdt and it will come out in the wash that they were and you will end up with a zero vector (when?)
Locoism said:
Also, how come Wperp will only have 2 vectors?
how many vectors span $\mathbb{C}^5$
Locoism said:
I think I'm missing what you mean when you say use the dot product. There are 6 possible combinations if Wperp has 2 vectors, and each one would equal zero. But wouldn't it take ridiculously long to write them all down and work out each one?

its not that bad - you could write the vectors as rows in a matrix, say {w_1,w_2,w_3} span W, and u is a vector in W perp
$$\begin{pmatrix} \textbf{w}_1^{T} \\ \textbf{w}_2^{T} \\ \textbf{w}_3^{T} \end{pmatrix} \textbf{u} = \textbf{0}$$

Hope this helps, but at the end of the day you have to try some work and ideas out

lanedance said:
you should probably understand how to check for linear independence prior to using gram-schmidt

Actually you can still attempt gram schimdt and it will come out in the wash that they were and you will end up with a zero vector (when?)
Ok I just didn't understand since I had already applied GS (and not gotten zero)
lanedance said:
how many vectors span $\mathbb{C}^5$
So Wperp should be the span of the "remaining" vectors? But why could it not be another set of 3 linearly independent vectors?

lanedance said:
its not that bad - you could write the vectors as rows in a matrix, say {w_1,w_2,w_3} span W, and u is a vector in W perp
$$\begin{pmatrix} \textbf{w}_1^{T} \\ \textbf{w}_2^{T} \\ \textbf{w}_3^{T} \end{pmatrix} \textbf{u} = \textbf{0}$$
Should those w's be the conjugates? (since we are using the standard inner product). I'm not too sure what you're getting at. I know the product should be zero, but how am I supposed to do that algebraically? Do I need to multiply that w matrix by an arbitrary vector an+bni where n is 1 to 5, and then solve? How would I end up with two different sets?

Locoism said:
Ok I just didn't understand since I had already applied GS (and not gotten zero)

So Wperp should be the span of the "remaining" vectors? But why could it not be another set of 3 linearly independent vectors?
wouldn't you then have 6 linearly independent vectors?
Locoism said:
Should those w's be the conjugates? (since we are using the standard inner product). I'm not too sure what you're getting at. I know the product should be zero, but how am I supposed to do that algebraically? Do I need to multiply that w matrix by an arbitrary vector an+bni where n is 1 to 5, and then solve? How would I end up with two different sets?
yes, it should be conjugates

write u=(a,b,c,d,e), then you have 5 unknowns whilst the matrix gives you 3 constraints, leading to a subspace spanned by 2 vectors - if you try some of these things it should become obvious

Oh yea.
I was going to do that for u, but shouldn't each a, b, c, d, e be in a complex form? In that case I would have 10 unknowns, and that just wouldn't work.
Also, I'm checking and re-checking my work, but I've ended up with
v1 = [1, 1+i, 1, 1-i, 1]
v2 = [2+i, -2+2i, -2+i, 2+2i, 2i]
v3 = (1/15)[-28+36i, 61-27i, -80i, -27-61i, -28-44i]
After apply G-S, which should give me an orthogonal basis, except these vectors are not orthogonal. I've checked it so many times, why are they not orthogonal?

are you normalizing your vectors first? the norm of |v1|>1

## 1. What is an inner product space?

An inner product space is a mathematical concept that refers to a vector space in which there is a defined notion of length and angle between vectors. It is equipped with an operation called an inner product, which takes two vectors as inputs and produces a scalar as an output. The inner product satisfies certain properties, such as linearity and positive-definiteness, and allows us to define the length of a vector and the angle between two vectors.

## 2. What is an orthonormal basis?

An orthonormal basis is a set of vectors in an inner product space that are all unit vectors (i.e. have a length of 1) and are all mutually orthogonal (i.e. have an angle of 90 degrees between any two vectors). This means that the basis vectors are independent and span the entire space, making it easy to represent any vector in the space using linear combinations of the basis vectors.

## 3. How do you find an orthonormal basis for a given inner product space?

To find an orthonormal basis for an inner product space, we can use the Gram-Schmidt process. This involves taking a set of linearly independent vectors and applying a series of orthogonalization and normalization steps to transform them into an orthonormal set. Alternatively, we can also use the eigenvalue-eigenvector method to find an orthonormal basis if the inner product space is defined by a symmetric matrix.

## 4. Why is an orthonormal basis important?

An orthonormal basis is important because it simplifies the representation of vectors in an inner product space. By expressing a vector in terms of the orthonormal basis vectors, we can easily calculate the length and angle of the vector, as well as perform other operations such as projections and rotations. It also allows us to easily compare vectors and perform calculations without worrying about the choice of basis.

## 5. Can an inner product space have multiple orthonormal bases?

Yes, an inner product space can have multiple orthonormal bases. This is because there are often many different ways to construct an orthonormal basis for a given space. However, all orthonormal bases for the same space will have the same number of vectors and will span the same subspace.