# Proof of subspace of R3

• robierob12

#### robierob12

Q:
Is the subset a subspace of R3? If so, then prove it. If not, then give a reason why it is not. The vectors (b1, b2, b3) that satisfy b3- b2 + 3B1 = 0

-----------------------
My notation of a letter with a number to the right, (b1) represents b sub 1.

Im having a problem on how far I need to go to show this is a subspace.

I know that since the equation that I am given is homogenous, the vector
(0,0,0) is included and it passes through the origin.

There is an example in my book where they say that because the (0,0,0) is not included it is enough to show it is not a subspace.

I know the test if u and v are vectors and
u and v are in W then u + v is in W
and if u is in W and c is any scalar, then cu is in W.

Do I need to use this test? or is the origin enough to be proof?

If the test is needed can someone bumb me in the direction of where to start this proof?

Take two vectors x and y from the set, and two scalars $\alpha$ and $\beta$. The given set is a subspace of R^3 if $\alpha x + \beta y$ is in the set.

Q:
Is the subset a subspace of R3? If so, then prove it. If not, then give a reason why it is not. The vectors (b1, b2, b3) that satisfy b3- b2 + 3B1 = 0

-----------------------
My notation of a letter with a number to the right, (b1) represents b sub 1.

Im having a problem on how far I need to go to show this is a subspace.

I know that since the equation that I am given is homogenous, the vector
(0,0,0) is included and it passes through the origin.

There is an example in my book where they say that because the (0,0,0) is not included it is enough to show it is not a subspace.

I know the test if u and v are vectors and
u and v are in W then u + v is in W
and if u is in W and c is any scalar, then cu is in W.

Do I need to use this test? or is the origin enough to be proof?

If the test is needed can someone bumb me in the direction of where to start this proof?

Yes, you need the test. radou has shown a quicker method but you will need to show it is non-empty. It is never implied.

The first step is to check if (0,0,0) is in there. That's the easiest way to show it's non-empty.

Second, if you're really terrible at checking if x + y is in the subset when x and y are in the subset, then just take two random vectors that is in the subset and add them. So, then you visually see what's going on.

b3- b2 + 3B1 = 0
For examples 2 of the following vectors is in the subset; you figure it out:

(9,3,-2)
(1,1,1)
(0,3,1)
(2,1,3)

Now, take the two that is in the set, and then add them. Is that new vector in the set?

After doing this, try showing for the general case for any two vectors in the subset.

Use the same approach to show it is closed under scalar multiplication.

so, I see what you we getting at and found the two vectors you were you were talking about added them and they were still in there. Even though you wrote the points in the form (b3, b2, b1)

For my proof then I re aranged the equation so it looks nicer

3b1 - b2 + b3 = 0

Let x and y be vectors in the subset. Is (x + y) also in the subset then?

3x1 - x2 + x3 = 0
3y1 - y2 + y3= 0

3(x1 + y1) - (x2+y2) + (x3 + y3) =

(3x1 + 3y1) + (-x2 - y2) + (x3 + y3) =

(3x1 - x2 + x3) + (3y1 - y2 + y3) =

0 + 0 = 0 by substitution of letting x and y be in the subset.

Will this cut it for Closure under addition?

Thanks for all the help so far guys... much appreciated.

Yes, it's okay. Note that, as I suggested earlier, you could have taken Ax + By, where A and B are scalars, and shown that it's closed both under addition and scalar multiplication in pretty much the same manner.

That's pretty sweet. I am going to try that... I didnt realize at first that would work for both. So even though My theorm test for a sub space shows, that there are two steps, they can be combined? nice. thanks

so, I see what you we getting at and found the two vectors you were you were talking about added them and they were still in there. Even though you wrote the points in the form (b3, b2, b1)

For my proof then I re aranged the equation so it looks nicer

3b1 - b2 + b3 = 0

Let x and y be vectors in the subset. Is (x + y) also in the subset then?

3x1 - x2 + x3 = 0
3y1 - y2 + y3= 0

3(x1 + y1) - (x2+y2) + (x3 + y3) =

(3x1 + 3y1) + (-x2 - y2) + (x3 + y3) =

(3x1 - x2 + x3) + (3y1 - y2 + y3) =

0 + 0 = 0 by substitution of letting x and y be in the subset.

Will this cut it for Closure under addition?

Thanks for all the help so far guys... much appreciated.

Looks good to me. But explictly write that (y1,y2,y3) and (x1,x2,x3) are two vectors in the subspace to be added.

That's pretty sweet. I am going to try that... I didnt realize at first that would work for both. So even though My theorm test for a sub space shows, that there are two steps, they can be combined? nice. thanks

The "steps" can be combined, since one can easily prove (you could try that, too) that the following two conditions for "being a subspace" are equivalent (if V is a vector space over a field F, and M a non-empty candidate for a subspace of V):

(1) for every x, y in M, x + y is in M & for every x in M and A in F, Ax is in M

(2) for every x, y in M and for every A, B in F, Ax + By is in M.

That's pretty sweet. I am going to try that... I didnt realize at first that would work for both. So even though My theorm test for a sub space shows, that there are two steps, they can be combined? nice. thanks

Yes, you can prove that they are equivalent.

But I warn you before using that, some TA's probably don't know that it is equivalent and what not. I generally avoid using that route unless a professor is marking it. The last thing I was is to lose marks and have to get them back afterwards. Trouble for nothing.

Thanks for the help on this one guys.

Then applied linear algebra class I am taking this semester is done with mostly all proofs, and this place has been a huge help for me when I get hung up on my work. Being that not many people in my class are into meeting up afterwards for disscusion, its nice to have a place where people are interested in the topic.

Rob