# How check for subspaces in Linear Algebra?

• saintdick
In summary, the given set is a subspace of R^4 if it satisfies closure under addition and scalar multiplication. This can be checked using algebraic theorems rather than randomly choosing values for x and y. To find a basis, we can rewrite any member of the set as a linear combination of two vectors, (2, 1, 0, -1) and (3, 0, 0, 2). Both of these vectors form a basis for the set.
saintdick

## Homework Statement

consider R^4. Let V be the set of vectors in the form ( 2x+3y, x, 0 , -x+2x) is this a subspace

of R^4 and why? find a basis if it's possible

## The Attempt at a Solution

I know that the set must work under scalar multiplication and vector addition but i am confused here because of the variables. how do you actually test that? do you just choose random values for x and y?

for the second part, i know that a basis could be found if it's a subspace but i am not sure how to find it?

IF you believe 'closure' of addition and scalar multiplication is false, yes, you could look for values of x and y which make the statement untrue. That is, look for a "counter-example".

However, if it is really a subspace, closure of addition and scalar multiplication must be true for all x and no matter examples you tried, that would never prove it true for all.

Instead use algebraic theorems to show the set is closed under addition and scalar multiplication. For this paraticular example, two "generic" members of the set would be of the form (2x+ 3y, x, 0, -x+ 2y) and (2x'+ 3y', x', 0, -x'+ 2y'). The sum of those is ((2x+ 3y)+(2x'+ 3y'), x+ x', 0+ 0, (-x+ 2y)+ (-x'+ 2y'))= (2(x+x')+ 3(y+ y'), x+x', 0, -(x+ x')+ 2(y+ y')). Do you see that this is in the same subset? It must be of the form (2x"+ 3y", x", 0, -x"+ 2y"). What are x" and y" in this case?

To see that it is closed under scalar multiplication, take a general member of the set, (2x+ 3y, x, 0, -x+ 2y) and multiply by the scalar "r":
(r(2x+ 3y), rx, r0, r(-x+ 2y))= (2(rx)+ 3(ry), rx, 0, -(rx)+ 2(ry)).
Again, do you see that is of the form (2x"+ 3y", x", 0, -x"+ 2y")? What are x" and y" here?

To find a basis, note that any member of this set is of the form (2x+ 3y, x, 0, -x+ 2y)= (2x, x, 0, -x)+ (3y, 0, 0, 2y)= x(2, 1, 0, -1)+ y(3, 0, 0, 2).

Thanks for your reply, but i am still confused? what to do you mean when you ask " What are x" and y" in this case". So let me get this straight, when you prove it the way you did, as long as the form doesn't change then that means it's a subset. For the basis, are both of those columns form one basis or do each one form a single basis?

## 1. What is a subspace in Linear Algebra?

A subspace in Linear Algebra is a subset of a vector space that satisfies three conditions: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication. In other words, if you add two vectors in the subspace together, the result is also in the subspace, and if you multiply a vector in the subspace by a scalar, the result is also in the subspace.

## 2. How do you check if a set is a subspace in Linear Algebra?

To check if a set is a subspace in Linear Algebra, you need to verify that it satisfies the three conditions mentioned above: containing the zero vector, being closed under addition, and being closed under scalar multiplication. You can also use the subspace test, which states that a set is a subspace if and only if all linear combinations of its vectors are also in the set.

## 3. What is the importance of subspaces in Linear Algebra?

Subspaces are important in Linear Algebra because they help us understand the structure of vector spaces and their properties. They also allow us to simplify calculations by breaking down a vector into smaller components that are easier to work with. Subspaces are also used in many applications, such as in computer graphics, data analysis, and physics.

## 4. Can a subspace contain only one vector?

Yes, a subspace can contain only one vector if that vector is the zero vector. This is because the zero vector satisfies all the conditions of a subspace: any vector added to it results in the same vector, and any scalar multiple of it is still the zero vector. However, if the subspace contains any other vector, it must also contain all scalar multiples of that vector to satisfy the closure under scalar multiplication condition.

## 5. How do you prove that a set is not a subspace in Linear Algebra?

To prove that a set is not a subspace in Linear Algebra, you need to find a counterexample that violates one of the three conditions: not containing the zero vector, not being closed under addition, or not being closed under scalar multiplication. You can also use the subspace test and show that there exists at least one linear combination of the vectors in the set that is not in the set.

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