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 \(\displaystyle \alpha= ax+by\).