Let V be an inner product space and 𝐯 ∈V. Prove that the set of all vectors 𝐰 ∈V that are orthog

Let V be an inner product space and 𝐯 ∈V. Prove that the set...

Key Concepts

Closure Properties

Closure under addition and scalar multiplication is a fundamental requirement for any subset to qualify as a subspace. This means that if any two vectors within the subset are added together, their sum must also be in the subset; likewise, multiplying any vector in the subset by a scalar must yield a vector that remains in the subset.

Subspace

A subspace is a subset of a vector space that is itself a vector space under the same operations of addition and scalar multiplication. In order to be a subspace, the set must contain the zero vector and be closed under vector addition and scalar multiplication, ensuring that all vector space axioms are inherited.

Inner Product Space

An inner product space is a vector space equipped with an inner product, a function that assigns a scalar to each pair of vectors. This structure allows for the concepts of length, angle, and orthogonality, making the space richer than a mere vector space and enabling geometric interpretations of algebraic operations.

Orthogonality

Orthogonality is the condition where two vectors have an inner product of zero, meaning they are perpendicular to each other with respect to the given inner product. This concept is central to many areas of mathematics and is used to decompose spaces, simplify computations, and understand projections.

Transcript

0:00 Hello there.

00:01 So for this exercise we're working with an inner product space v that have a basis b sub b defined by the vectors b1, b2 up to vr.

00:14 And we need to prove that the only vector that is orthogonal to this space b is the zero vector.

00:20 So for that let's first consider so let's consider a subspace of b defined as the set of all the vectors w in this in the in the inner product space b such that the inner product of w with v i is equal to zero for all the v i's in the basis of b so we need to show that this subset has uh is a is a only only contains the zero vector okay so basically is to say that w is a singlet so let's show this so first we need to show that this set is non empty and we know by definition of the inner product that the zero vector the inner product of the zero vector with any vector b is equals to zero for every v on the on the space so this is the general definition of of the inner property of the inner product and this implies that then the zero vector is also equals to zero the inner product of the zero vector with any vector in the basis it's going to be equals to zero for every vi in the basis so we can say that the zero vector is an element on w and this implies that w is not which is great but then we need to show that w has an as an only element the dog the zero vector so for that we need to consider a generic vector in in the subspace w so let's consider so any vector in w due to w is a subspace of b is a equivalent to say that w can be written as a linear combination from i up to r of the basis elements okay because if w is an element of the subspace w then w is also an element of the vector space and if w is an element of the vector space b then this can be expressed as a linear combination of the basis elements so then we can consider the inner product of w with itself.

03:34 And we are going to express this w as a linear combination of the basis elements of b...

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