Proof of Subspace: x-y-3z=0 in R^3

[pokgai]

1. prove whether x-y-3z=0 is a subspace of R^3 or not

Homework Equations

for proofs
1. set must not be empty
2. set is closed under vector addition
3. set is closed under scalar multiplication

The Attempt at a Solution

Not sure if this is correct, but what I did was find the vector equation of the x-y-3z=0
which equals (x,y,z) = (0,0,0) +s(1,1,0)+t(3,0,1)
i defined (1,1,0) as v and (3,0,1) as u
so i added v+u and it remains true in that it equals 0
and then the scalar i just used any real value and either v or u and found that also equals 0
so hence it was a subset of R^3.. and also that (0,0,0) was defined..


not sure if that's what you're meant to do but yeah does that sound right?

 

[praharmitra]

your basic plan of action is correct. But, just to frame it in a better (and clearer) way, write

\vec{r}(s,t) = (x,y,z) = \vec{0}+s\vec{u}+t\vec{v}

So any point on this space is determined by specifying s and t.

1. Show that it is not empty (Set some particular values of s and t, and show that there is atleast one point.

2. Closed under vector addition - Show that for a particular choice of (s1,t1) and (s2,t2), there always exists (s3,t3) such that \vec{r}(s_1,t_1)+\vec{r}(s_2,t_2)=\vec{r}(s_3,t_3)

4. Closed under scalar multiplication - Show that for every choice of (s,t) and 'a' there exists a (s',t') such that a\vec{r}(s,t) = \vec{r}(s',t')

Though you have already done all of the above, you should frame it as shown.

 

[HallsofIvy]

1. prove whether x-y-3z=0 is a subspace of R^3 or not

Homework Equations

for proofs
1. set must not be empty
2. set is closed under vector addition
3. set is closed under scalar multiplication

The Attempt at a Solution

Not sure if this is correct, but what I did was find the vector equation of the x-y-3z=0
which equals (x,y,z) = (0,0,0) +s(1,1,0)+t(3,0,1)
i defined (1,1,0) as v and (3,0,1) as u
so i added v+u and it remains true in that it equals 0

Be careful with your grammar! You used "it" to mean two different things in this sentence. I think what you are saying is that v+ u also satisfies the equation x- y- 3z= 0 and so is in the set.
Since every vector in the set is of the form su+ tv for some numbers a and b, and the set is closed under scalar multiplication, yes, that is sufficient- except that you haven't yet proved the set is closed under scalar multiplication.

and then the scalar i just used any real value and either v or u and found that also equals 0

Again with the "it"! Use fewer pronouns and you will be clearer- you mean that ku and kv still satisfied the equation. And, you really needed to put this before the the proof of closure of addition.

so hence it was a subset of R^3.. and also that (0,0,0) was defined..

You were given that this is a subset! You mean it was a subspace. Of course, you can't say that until after you said it was non-empty. Finally, you don't mean to say that (0, 0, 0) was "defined"- of course the 0 vector is defined the question is whether or not it is in the subset. And you should specifically point out that it is because 0- 0- 3(0)= 0.

not sure if that's what you're meant to do but yeah does that sound right?