# linear space and subspaces

• May 18th 2010, 02:03 AM
jayshizwiz
linear space and subspaces
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 $\displaystyle R^4$.
• May 18th 2010, 04:23 AM
HallsofIvy
Quote:

Originally Posted by 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 $\displaystyle R^4$.

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?
• May 18th 2010, 07:06 AM
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...
• May 18th 2010, 08:39 AM
HallsofIvy
Quote:

Originally Posted by 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...

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.

Quote:

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 $\displaystyle \alpha= ax+by$.
• May 19th 2010, 12:44 PM
jayshizwiz
Thank you so much!