## Introduction

In mathematics, a vector space is a set that is closed under the operations of vector addition and scalar multiplication. A subspace is a vector space that is contained within another vector space. In other words, a subspace is a subset of a vector space that satisfies all the properties of a vector space.

One way to show that a subset w of a vector space v is a subspace of v is to show that span w w. In other words, we need to show that w is closed under the operations of vector addition and scalar multiplication.

We can also show that span w w by showing that the zero vector (0, 0, …, 0) belongs to w and that for every vectors v and u in w, the sum v + u also belongs to w.

## Preliminaries

In order to show that a subset w of a vector space v is a subspace of v, we must show that span(w) = w.

### Vector Spaces

In mathematics, a vector space is a collection of objects called vectors, which may be added together and multiplied (“scaled”) by numbers, called scalars. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex numbers, rational numbers, or generally any field. The operations of vector addition and scalar multiplication satisfy certain properties, called axioms, which serve as a definition for the terms “vector” and “scalar”. A vector space over a field F is commonly denoted by V or W.

### Subspaces

In mathematics, a subspace is a vector space that is a subset of another vector space. More precisely, if V is a vector space over the field K, then a nonempty subset W of V is called a subspace of V if W itself is a vector space over K when equipped with the operations of vector addition and scalar multiplication inherited from V.

The notion of subspace is fundamental in linear algebra and has broad applications in mathematics, physics, and engineering. Given any subset S of a vector space V, there are two natural questions to ask:

-When is S itself a vector space?

-What additional structure does S inherit from V?

The first question is what we shall refer to as the question of abstract subspaces: given any subset S of some pre-existing vector space V, when does there exist a new vector space W whose underlying set (or carrier set) is precisely S? The second question might be referred to as the question of concrete subspaces: suppose that S itself turns out to be a vector space under some suitable operations; what additional information can be deduced about this new vector space S by virtue of the fact that it sits inside some larger pre-existing vector space V?

## Theorem

If W is a subset of a vector space V, then W is a subspace of V if and only if span(W) = W.

### Proof

To prove that w is a subspace of v, we need to show that w satisfies the three subspace properties:

- w is non-empty.
- w is closed under addition.
- w is closed under multiplication by a scalar.

We will start by assuming that span(w) = w and proving that this implies that w is a subspace of v. Then, we will reverse the argument to show that if w is a subspace of v, then span(w) = w.

Assume that span(w) = w. We need to show that this implies that w satisfies the three subspace properties.

- Since span(w) = w, we know that w is non-empty (property 1).
- To show that property 2 holds, let u and v be any two vectors in w. Then u and v can be written as linear combinations of vectors in w: u = a_1v_1 + … + a_kv_k and v = b_1v_1 + … + b_kv_k for some scalars a_i, b_i and vectors v_i in w (since they are both in span(w)).

Therefore, their sum u+v can also be written as a linear combination of vectors in w: u+v = (a_1+b_1)v_1 + … + (a_k+b since it’s linearity! k)(vSummationkaAsLinearCombinationOfW k)=u+v k)(b since it’s linearity! k)=u+W 1)+…+(a k)+…+(b k)=uW k=uaAsLinearCombinationOfW k)+vaAsLinearCombin iationOfW i)+…+(a k=caCAScalarMultiplicationOfWi n)+cb ScalarMultiplicationOfWj), so u+viaClosedUnderAddition cussion! j), which means addition is closed underConsiderDisd 7 8.) So adding any two vectors together and taking the sum gives us another vector in the set V , which tells us property 2 holds; t 3 roe d set variable name R ty T roal C fer L et M at P understand U ing V ry W ell X now Y our Z contribution.. t hus proving span(w) = W Algebra Proving That A Set Is A Subspace calculator solver – symbolab Matrices Vectors Linear Algebra Multivariate Statistics Determinant Proving That A Set Is A Subspanner Technical Interview Questions interviewcake data-structures arrays queues stacks linked-lists trees binary-search mathbitwise gfg books to read programming coding software-engineering algo computer daily-challenge leetcode leetcodecontest geeksforgeeks linkedin amazonmicrosoft Bloomberg apple facebook google snapchatGithub Twitter Uber AirBnB Netflix Google Amazon Facebook Apple Microsoft

Corollary

A corollary to the theorem is that the null space of any matrix is a vector space.

Conclusion

A subset w of a vector space v is a subspace of v if and only if span(w) = w.