# Linear algebra, subspace.

1. Mar 3, 2008

### frasifrasi

The question I am looking at asks,

Let a be the vector (-1,2) in R^2. Is the set S = { x is in R^2 | x dot a = 0}
a subspace?

--> x and a are vectors...

Can anyone explain how to show this?

I was thinking that since the zero vector is in R^2, this must also be a subspace.

Thank you.

2. Mar 3, 2008

### quasar987

Since 0 is in S, you can't conclude that S that S is not a subspace.

Is it a subspace? To aswer this, you must show that S is closed under addition and scalar multiplication. It is pretty direct, try it.

3. Mar 3, 2008

### frasifrasi

Oh yes.
Ok, but how would I proceed.

Do i just look at 0 -- k(0) = (k0).

I am not sure how to apply the axiums, can you/someone demonstrate?

Thank you.

4. Mar 3, 2008

### quasar987

I'll do closure under addition for you.

Let x and y be in S. We want to show that x+y is in S also.

well (x+y) dot a = (x dot a) + (y dot a) = 0 + 0 = 0,

so x+y is in S!

(I used the distributivity property of the scalar product)

5. Mar 3, 2008

### frasifrasi

For the constant,

k(x dot a) = (kx dot a)
k(0) = 0
0 = 0

Is this the proper way?
Thanks for the help.

6. Mar 3, 2008

### quasar987

The exposition is not very logical.

Try again using my template.

"Let x be in S and k be a constant. We want to show...."

7. Mar 4, 2008

### HallsofIvy

Why do you keep looking at 0? Not only does just proving that 0 is in the set not prove it is a subspace, you don't need to prove that 0 is in the set!

You need to prove two things: the set is closed under addition and the set is closed under scalar multiplication.

Suppose u and v are two vectors such that $u\cdot a= 0$ and $v\cdot a= 0$ where a is a fixed vector. Can you prove that u+ v is also in that set- that $(u+v)\cdot a= 0$?

Suppose u is a vector such that $u\cdot a= 0[/tex] and [itex]\alpha$ is any real number. Can you prove that $\alpha$ is also in that set- that $\alpha u\cdot a= 0$?

After you have proved the second of those, you don't need to prove 0 is in the set- it follows from the fact that 0v= 0 for any v.

8. Mar 4, 2008

### frasifrasi

so, au dot a = 0

u dot a = 0
so, u(0) = 0.

Is that enough?

thank you for the patience everyone.

9. Mar 4, 2008

### HallsofIvy

I have no idea what you mean by this. You started with a a vector so I don't know what you mean by "au dot a= 0". Did you accidently also use a to mean a number?

What do you mean by u(0)? u is a vector, not a function.

If I try to translate, I get "If ku dot a= 0, then u dot a= 0".
That's exactly the opposite of what I suggested: Show that if u dot a= 0, then ku dot a= 0 for any real number k.

And, of course, you must prove that if u dot a= 0 and u dot v= 0, then (u+ v) dot a= 0.

10. Mar 4, 2008

### frasifrasi

like the poster above mentioned,

(u+v) dot a = 0
then

(u dot a) + (v dot a) = 0
0 + 0 = 0 so, that is closed under addition.

For scalar multiplication, yes I meant k:

k(u dot a) = k.0 = 0

ku dot a = k.0 = 0

Is this the right way to prove it?

11. Mar 4, 2008

### quasar987

Maybe you understand, but I wouldn't know because your exposition lacks words or at least "<==>" symbols for it to make any sense.

In your proof of scalar multiplication, the critical part of the proof is the fact that ku dot a = k(u dot a). I see you nowhere make that connection explicitely.