I am having trouble understanding linear space and subspace. For example:

Given that set and

Prove that k is not a subspace of .

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- May 18th 2010, 02:03 AMjayshizwizlinear space and subspaces
I am having trouble understanding linear space and subspace. For example:

Given that set and

Prove that k is not a subspace of . - May 18th 2010, 04:23 AMHallsofIvy
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 AMjayshizwiz
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 , -5 + 0 = -5 and so on...

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

W = 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 AMHallsofIvy
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 = 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...

**every**vector of that form - for**every**real number - is in that set. (5, -5, 0, 0) is in the set, taking , but so is (10, -10, 0, 0), taking , (25, -25, 0, 0), taking , and so is the 0 vector, (0, 0, 0, 0), taking .

Taking , (x, -x, 0, 0) is in the set and taking 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 . - May 19th 2010, 12:44 PMjayshizwiz
Thank you so much!