# How to prove a linear subspace

*A quick methodology to check that a set is a linear — or vector — subspace of a bigger vector space.*

# What is a linear subspace ?

To understand the idea of linear subspace you first have to got a good understanding of the concept of space and vector space. So what is a vector space?

## Vector space

Mathematics are infinite, you have an infinite amount of possibilities, of questions and of answer. A space is a defined part of this infinite. A vector space is a defined part of this infinite which respect three rules: (1) this space is made of vector ; (2) those vector are summable together ; (3) those vector are multipliable by a scalar.

## Different forms of vector space or set

In general, we can have two forms of representation of sets or vector space:

**The first form**, above, is structured as follows: the set A is the set of vectors *(x,y,z)* which respect the following properties: *x+y+z=0* and *x-2y=0*.

We find for example in this set the vector (2,1,-3), it has well the form (x,y,z) and respects well the two equality: *x+y+z=2+1–3=0 *and *x-2y=2–2×1=0*.

The idea is that this form of presentation of a set is arranged in the following way:

(1) we are presented with the general form of the set (with A for example it is a vector of size 3) and then ;

(2) we are given the characteristics that the elements of the set must respect (with A it is the two equations).

**The second form**, above, is structured as follows: the set B is the set, formed by the variables *x*, *y* and *z* belonging to *R*, of vectors of the form *(x+2y, y, z-3y+x)*.

For example, the set B would have the vector (3,1,-1) that we “generated” by taking *x=y=z=1* so we had *x+2y=1+2=3*, *y=1 *and *z-3y+x=1–3+1=-1*.

The idea is that this form of presentation of a set is arranged in the following way:

(1) we are given the variables to generate our set (for B it will be x,y,z) then ;

(2) we are given the “formula” to “generate” our set (for B it would be: (x+2y,y,z-3y+x)).

*Note: these two forms are representation forms, we could rewrite form 1 as form B, it is just a choice of presentation of the set.*

## So what is a subset ?

In linear algebra

# Proving a linear subspace — Methodology

*To help you get a better understanding of this methodology it will me incremented with a methodology.*

I want to prove that the set **A** is a linear sub space of **R³**.

## Proving that the set include the null space

First, we want to know if our set **A** contains the empty set of **R³**.

In mathematic, we’ll note :

To do that, we check that the vector defined by our set can be equal to the null vector of the same size.

If the vector defined by our set can be equal to the null vector then it means that our set **A** contains the empty set of **R³**. Now we have to validate the steps (2) and (3), stability by addition and then by product, to prove that the set **A **is indeed, or not, a sub-vector space.

If, on the contrary, the vector defined by our set cannot be equal to the null vector, then it means that our set A does not contain the empty set of R³. So we can finish the exercise here by concluding that our set is not a sub vector space of R³.

In our example, we can find a solution that is the following :

So indeed, our set **A** contains the empty set of **R³.**

## Proving the stability of the set by addition

**If step (1) is validated**

We now seek to prove that our set **A** is stable by addition. That is, if I take two vectors of my set and add them together, I get a vector of my set.

In mathematics we will note that we are trying to prove:

So we take two vectors, which we can name *u* and *v* from our set […] :

We sum them in a new vector, which we will name *w *:

We check now that the vector *w* is well part of the set A, that is to say that it respects well the properties of the set A.

avec :

If the vector *w* is part of the set A then we have proved that the set A is stable by addition.

If the vector *w* is not part of the set A then we have proved that the set A is not stable by addition and therefore is not a sub vector space of the set R³.

In our example, A is indeed stable by addition, because w is a vector of A.

## Proving the stability of the set by product with a scalar

**If step (2) is validated**

We now seek to prove that our set A is stable by multiplication. That is, if I take a vector of my set and multiply it by a scalar (a variable of the real set), I obtain a vector of my set.

In mathematics we will note that we are trying to prove:

So we take a vector from the set A, which we will name *u* :

We multiply the vector *u* by a scalar *λ* which gives us a new vector that we will name *w*:

We now check that the vector *w* is indeed part of the set A, that is to say that it respects the properties of the set A.

If the vector w is indeed part of the set A then we have proved that the set A is stable by multiplication. Finally, we can conclude that our set A is indeed a sub vector space of the set R³.

If the vector *w* is not part of the set A then we have proved that the set A is not stable by multiplication, and therefore is not a sub vector space of the set R³.

In our example, the set A is indeed stable per multiplication, and so A is a sub vector space of R³.