Show that the standard basis vectors 𝐞1, 𝐞2, 𝐞3 form an orthogonal basis with respect to the wei

Show that the standard basis vectors 𝐞1, 𝐞2, 𝐞3 form an...

Key Concepts

Weighted Inner Product

A weighted inner product is a generalization of the standard dot product in which different coordinates (or components) of the vectors are scaled by given weights. In this context, the inner product is defined so that each term in the product of the vector components is multiplied by a specific weight. This approach modifies the geometry of the space, determining lengths and angles differently than the usual Euclidean inner product.

Orthogonal Basis

An orthogonal basis is a set of vectors that are all mutually perpendicular with respect to a given inner product. This means the inner product of any two different basis vectors is zero. The property of orthogonality simplifies many computations in linear algebra, such as projections and decompositions, since the components of a vector along the basis vectors do not interact with each other.

Orthonormal Basis

An orthonormal basis is an orthogonal basis in which each vector has been normalized to have unit length with respect to the given inner product. In other words, each vector in the orthonormal basis satisfies that its inner product with itself is one. This normalization further simplifies computations, allowing for straightforward expansions of vectors as simple sums of scaled basis vectors.

Normalization Process

The normalization process involves taking each vector in an orthogonal basis and dividing it by its norm, computed using the weighted inner product. This process converts the orthogonal basis into an orthonormal basis while preserving the directions and the orthogonality between different vectors. It is essential when the aim is to work with unit vectors to simplify applications such as coordinate transformations and projections.

Transcript

00:01 In problem 7 we have these two vectors that is a basis for a subspace w we want to get an ortho -normal basis for these subspace the first step is to get v1 which equals one of these two vectors choose any one choose the first vector to minus 5 and 1 the second step is to get v2 from the project of the second vector here on the on v1 which equals the second vector 4 minus 1 2 minus you want to project onto v1 the rejection onto v1 we will write v1 here 2 minus 5 1 and we multiply by a factor this factor has in the nominator the dot product between the second vector 4 minus 1 2 and v1 2 minus 5 1 and in the denominator we have the dot product between v1 and v1 2 minus 5 1 dot 2 minus 5 1 it equals 4 minus 1 2 minus 1 2 minus the product is 4 multiplied 2 plus 1 multiple 2 plus 1 multiple by 5 plus 2 multiplied by 1 equals 15 the second product is 2 squared plus 5 squared plus 1 squared equals 30 multiplied by 2 minus 5 1 it equals 4 minus 1 which is 3 and we have minus 1 plus 2 .5 5 5 and we have minus 1 plus 2 .5 we have 1 finally we have 2 minus half multiplied by 1 which is 1 .5 this is the orthogonal basis to get the ortho normal basis we just get we just normalize v1 and v2 the third step is to normalize v1 and v2 to normalize v1 and get n1 for example equals we divide by its length we divide by the length of the vector 1 divided by the length of the vector is just the square root of its dot product the square root of v1 dot v1 we have its value here 30 square root of 30 multiplied by 2 minus 5 1 it equals 2 divided by square root of 30 divided by minus 5 divided by square root of 30 and finally 1 divided by square root of 30 the second vector the normalization will be the same 1 divided by its norm or its length which is the square root of v2 dot v2 square root of 4 squared plus 1 squared plus 2 squared plus 2 squared multiplied by v2 which is 4 minus 1 2 no this is not v2 this is x2 this is x2 we want to normalize v2 this here is v2 we want to normalize v2 square root of 3 squared plus 1 .5 squared plus 1 .5 squared multiplied by v2 which is 3 1 .5 1 .5 equals we have 1 divided by 1 .5 multiplied by square root of 6 multiplied by v2 3 1 .5 1 .5 1 .5 then equals 2 divided by square root of 6 and 1 divided by square root of 6 and finally 1 divided by square root of 6...

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