# linear space and subspaces

#### jayshizwiz

I am having trouble understanding linear space and subspace. For example:
Given that set $$\displaystyle k \subseteq R^4$$ and $$\displaystyle k = \{(5, -5, 0, 0)\}$$
Prove that k is not a subspace of $$\displaystyle R^4$$.

#### HallsofIvy

MHF Helper
I am having trouble understanding linear space and subspace. For example:
Given that set $$\displaystyle k \subseteq R^4$$ and $$\displaystyle k = \{(5, -5, 0, 0)\}$$
Prove that k is not a subspace of $$\displaystyle R^4$$.
You are saying that k contains the single vector v= (5, -5, 0, 0)? A subspace must be closed under addition. Is x+ x in the subspace?

Or- a subspace must be closed under scalar multiplication. Is 5x in k?

Or- a subspace must contain additive inverses. What is the additive inverse of x? Is it in k?

Or- a subspace must contain the 0 vector. Is the 0 vector in k?

#### jayshizwiz

I think that's where i got confused...There is only 1 vector. I was using the elements of the vector to prove that it is a subspace...

k = (5, -5, 0, 0)

I was saying that there is closure under addition: 5 + (-5) = 0 $$\displaystyle \subseteq k$$, -5 + 0 = -5 $$\displaystyle \subseteq k$$ and so on...

But then please explain why my book says this is a subspace of R^4:

W = $$\displaystyle \{(\alpha, -\alpha, 0, 0) | \alpha$$ is a real number}

Does this mean that W is also only a single vector or alpha and minus alpha contain four elements each?? This is where I'm getting very confused?? The book also says that W contains k...

#### HallsofIvy

MHF Helper
I think that's where i got confused...There is only 1 vector. I was using the elements of the vector to prove that it is a subspace...

k = (5, -5, 0, 0)

I was saying that there is closure under addition: 5 + (-5) = 0 $$\displaystyle \subseteq k$$, -5 + 0 = -5 $$\displaystyle \subseteq k$$ and so on...
No, it is not a matter of the individual components. x+ x= (5, -5, 0, 0)+ (5, -5, 0, 0)= (10, -10, 0, 0) which is NOT in the set.

5x= 5(5, -5, 0, 0)= (25, -25, 0, 0) which is NOT in the set.

And, of course, (0, 0, 0, 0) is NOT in the set.

But then please explain why my book says this is a subspace of R^4:

W = $$\displaystyle \{(\alpha, -\alpha, 0, 0) | \alpha$$ is a real number}

Does this mean that W is also only a single vector or alpha and minus alpha contain four elements each?? This is where I'm getting very confused?? The book also says that W contains k...
No, the notation $$\displaystyle \{(\alpha, -\alpha, 0, 0)|\alpha$$ is a real number} means that every vector of that form - for every real number $$\displaystyle \alpha$$- is in that set. (5, -5, 0, 0) is in the set, taking $$\displaystyle \alpha= 5$$, but so is (10, -10, 0, 0), taking $$\displaystyle \alpha= 10$$, (25, -25, 0, 0), taking $$\displaystyle \alpha= 25$$, and so is the 0 vector, (0, 0, 0, 0), taking $$\displaystyle \alpha= 0$$.

Taking $$\displaystyle \alpha= x$$, (x, -x, 0, 0) is in the set and taking $$\displaystyle \alpha= y$$ so is (y, -y, 0, 0). For any real numbers a and b, a(x, -x, 0, 0)+ b(y, -y, 0, 0)= (ax+ by, -ax- by, 0, 0)= (ax+by, -(ax+by), 0, 0) which is in the set taking $$\displaystyle \alpha= ax+by$$.

#### jayshizwiz

Thank you so much!