Is U a Valid Subspace of R3 When x Equals z?

[roam]

Homework Statement

Let U={(x,y,z) \in R³ : x=z}. Show that U is a subspace of R³.

Homework Equations

The Attempt at a Solution

U is non-empty it contains the 0 vector:

U= {(x,y,z) = (s,t,s), s,t \in R}

U={s(1,0,1)+t(0,1,0), s,t \in R}

for s,t=0
0(1,0,1)+0(0,1,0)=(0,0,0)

closed under addition:

u=(s1,t1,s1)
v=(s2,t2,s2)

u+v=(s1+s2,t1+t2,s1+s2)
Hence x=z

This was my attempt so far, I'm not sure how to prove the 3rd condition... can anyone show me how to prove that it is closed under scalar multipication?

Regards

 

[Focus]

I assume you are taking scalars in R.
\lambda (x,y,x)=?

 

[jambaugh]

You should take an arbitrary element of the set and multiply by an arbitrary scalar and then show the result is still in the set. That's what the definition means!

 

[roam]

So,

A vector u=(s1,t1,s1) belongs to the set.

for a scalar \lambda \in R

λ(s1,t1,s1)=λs1,λt1,λs1


Is this all I need to write??

 

[Mark44]

So,

A vector u=(s1,t1,s1) belongs to the set.

for a scalar \lambda \in R

λ(s1,t1,s1)=λs1,λt1,λs1


Is this all I need to write??

Does your vector u belong to set U? If so, you should have that as your conclusion.

Also, from your work in an earlier post, does your vector u + v belong to set U? If it does, you should say that. You concluded "hence x = z" which is not what you should conclude.