Is R={(x,y,z,w) belongs to R^4; x+y+sz-w=s^2-2s} a subspace of R^4?

[Neoon]

I came across a problem in linear algebra asking me to find for which values of s the following set:
R={(x,y,z,w) belongs to R^4; x+y+sz-w=s^2-2s}
is a subspace of R^4 with respect to the usual additoin and scalar multiplication. Any idea to solve this question?

Thanks in advance

 

[Office_Shredder]

Suppose $$(x_1,y_1,z_1,w_1)$$ and $$(x_2,y_2,z_2,w_2)$$ are in the set.

Then $$x_1+y_1+sz_1-w_1 = x_2+y_2+sz_2-w_2=s^2-2s$$

and

$$ax_1+ay_1+asz_1-aw_1=s^2-2s$$

For 'a' real.

This should tell you everything you need to know about what s can be actually... I would use the second one to start

(In case you didn't realize, this is saying "well, let's see if our set is closed under addition and scalar multiplication, since if it is, it satisfies the vector space axioms)

 

[Neoon]

Thanks for your quick response
What is the puropose of a in the second equation? i.e. Why you multiplied each term of the left hand side by a.
I appreciate if you give me more hints.

 

[Office_Shredder]

Because, if your set is a subspace, it must be closed under scalar multiplication. Which means if

$$(x_1,y_1,z_1,w_1)$$

is in your set, you want

$$a*(x_1,y_1,z_1,w_1)$$

to be in your set. But this is just

$$(a*x_1,a*y_1,a*z_1,a*w_1)$$

and hence, this quadruplet must satisfy the original condition of

$$x+y+sz-w=s^2-2s$$

i.e. $$(a*x_1)+(a*y_1)+s(a*z_1)-(a*w)=s^2-2s$$

for all $$a \in R$$

So try thinking of "good" values of a that yield useful information about what s might be

 

[Neoon]

Thank you for your help. I

 

[Neoon]

Sorry. I would like to tell you that I am a chemical engineer and I have a math course (mathematics for engineering). One chapter of that course is about eigenvalues and eigenvectors and our instructor refer only to the subspaces and it is not mentioned in the book. I tried last week to read many books about the subspaces to be able to solve the question but I was not able to understand the problem and half of knowing the answer is understanding the question.

Any way, How I will be able to make (a*x_1, a*y_1, a*z_1, a*w_1) in set?
What is the useful information that I should get?

Until now the problem is not understood to me?
I appreciate your help.

 

[Office_Shredder]

A bit of a crash course about subspaces then.

A vector space is a set of objects (called vectors, naturally) that satisfy some axioms (you can read about them at
http://en.wikipedia.org/wiki/Vector_space) .

Basically, it just says that addition and multiplication work how you would want them to (associativity, distribution, etc.)So a subspace is defined as a subset of your vector space, that is in itself a vector space.

So let's say we have a vector space V. We also have a subset U, and want to know if U is a subspace (aka, show if U is a vector space). If U is closed under addition and scalar multiplication (so if u₁, u₂ are in U, and a,b are real numbers, a*u₁ + b*u₂ are in U), then the rest of the vector space axioms hold (because they must hold in V, and U is a part of V), and thus U is a subspace.

So we test this... call U the subspace of (x,y,z,w) such that x+y+sz-w=s^2-2s. We're assuming it's a subspace, so we need to find, given what we know about subspaces, what s can be. We know subspaces are closed under multiplication of scalars, so if (x,y,z,w) is in U, so is a*(x,y,z,w) = (ax,ay,az,aw) and thus, plugging that quadruplet right back into the definition of what's in our subspace
ax+ay+saz-aw = s²-2s

This is true for any real number a.

So, this means, for example, that if a=2, this must be true. I.e. 2x+2y+s2z-2w = s²-2s
Factoring 2
2(x+y+sz-w) = s²=2s

But we originally had x+y+sz-w = s²=2s

So what can we conclude?

 

[Neoon]

let us see if i am understanding. ax+ay++saz-aw=s^2-2s is closed under multiplication if and only if the RHS of the equation s^2-2s =0
s^2=2s
divide by s
s=2

So, if s=2 the vector space is closed under multipliaction

in other words,
ax+ay+saz-aw=x+y+sz-w = s2-2s if and only if s=2.

Is this correct?

 

[Office_Shredder]

you're close. There is one other value of s that you missed (by dividing it out...).

Once you have the values of s such that the subspace is closed under multiplication, you need to show that it's closed under addition too. It can be a little tricky the first time if you don't know what exactly you're trying to do, but really isn't that difficult. So, given your value of s (which makes the RHS zero),

show if $$(x_1,y_1,z_1,w_1)$$ and $$(x_2,y_2,z_2,w_2)$$ satisfy x+y+sz-w = 0, then so does $$(x_1+x_2, y_1+y_2, z_1+z_2, w_1+w_2$$

 

[Neoon]

May be the the trivial solution s=0

Now, let's say
If u= (x1,y1,z1,w1) is in R then x1+y1+sz1-w= s^2- 2s for some number s. If v= (x2,y2,z2,w2) is in R then x2+y2+tz2-w= t^2-2t for some number t.

if s=0 or 2 and t=0 or 2 then au+ bv= (ax1+bx2,ay1+ by2,az1+bz2,aw1+bw2) which is satisfy the equation x+y+sz-w=s^2-2s

then it is close under addition. the values of s= (0,2)

Am I reaching the answer of the problem?

if yes, what about if
R={(x,y,z,w) belongs to R^4; x+y+z-w=s^2-2s}
or
R={(x,y,z,w) belongs to R^4; x+y+sz-sw=s^2-2s}
or
R={(x,y) belongs to R^2; x+y=s^2-2s}
or
R={(x,y,z,w) belongs to R^2; x+y+sz-sw=s^2-2s} (is this statement true)

Is the answer the same ?

what about the third condition of R to be called a suspace that is R should conatin a zero vector in R^4.

?

 

[Office_Shredder]

As long as you're closed under addition and multiplication, you always have the zero vector in your set!

First, in all your examples, as long as the RHS is 0, the zero vector trivially solves it. If it doesn't though:

Suppose v belongs to your (possible) subspace. Then -1*v belongs also, because we have shown it's closed under scalar multiplication. Hence, v+(-v) is in it also. But this is the zero vector, and we're done.

And yes, all those examples you posted require s to be 0 or 2. Note though, that they aren't the same vector space.

I don't think you quite wrote enough to show it's closed under addition. You'll want to do something like this:

Suppose $$(x_1, y_1, z_1, w_1)$$ and $$(x_2, y_2, z_2, w_2)$$ satisfy $$x+y+sz-w = 0$$ since we determined $$s^2-2s = 0$$

Then we want to show $$(x_1+x_2) + (y_1+y_2) + s(z_1 + z_2) - (w_1 + w_2) = 0$$

Rearranging gives us

$$[x_1 + y_1 + sz_1 - w_1] + [x_2 + y_2 + sz_2 - w_2]$$

and what can yo usaw about the term in each pair of brackets?

 

[Neoon]

each bracket is equal to zero
[x1 + y1 + sz1 - w1] + [x2 + y2 + sz2 - w2]= 0+0=[S^2-2s]+[t^2-2s]
for t=s=(0,2)

 

[Office_Shredder]

And that about sums it up.

Congratulations, you're an expert on subspaces

 

[Neoon]

Thank you. I won’t be without you help. I spent last week reading on subspaces but I wasn’t able to understand it while the idea was not diffuclt.

Now, let’s add this to the problem. “Find its dimensions and provid a basis for it”

 

[Office_Shredder]

Now it depends on what value of s you have (0 or 2).

First, do you know what the definition of a basis and dimension are?

 

[Neoon]

Yes I have read some books that mintion the definition of the basis and dimension.
Basis: is a linearly independent set of vectors that spans the space. My problem is with the span because for example the following three vectors
(1 (0 (0
0 1 0
0) 0) 1)
spans the vector space R^3. but I can generate a lot of vectors form these three vectors which changes the basis. I am really confused in this part.

for the dimension: it is n if and only if V has a basis containing n vectors, where V can not be zero. As I understood, I should find the basis to find the dimension.

In addition, in some examples I saw some books saying the dimension, for example, is 2 if R^2 and 4 if R^4. Does this apply also here. Can I say the dimension is 4 in our case because we have R^4? And what changes if s=0 or s=2?

I really appreciate your help

 

[radou]

but I can generate a lot of vectors form these three vectors which changes the basis. I am really confused in this part.

What do you mean? The basis isn't unique; any linearly independent set that spans the space is a basis.

for the dimension: it is n if and only if V has a basis containing n vectors, where V can not be zero. As I understood, I should find the basis to find the dimension.

Yes, you should.

In addition, in some examples I saw some books saying the dimension, for example, is 2 if R^2 and 4 if R^4. Does this apply also here. Can I say the dimension is 4 in our case because we have R^4? And what changes if s=0 or s=2?

I really appreciate your help

If you're looking at R^n, the dimension is n. If you're looking at a subspace of R^n, the dimension is less or equal to n.

 

[Neoon]

Thanks Radou

You said ("Yes, you should" find the basis to find the dimension). But the question statement is as the following:" Find the dimension of the subspace and provide a basis for it", so here it is the other way around. I am really confused. Could you please clarify this point for me?

 

[Office_Shredder]

I think it's supposed to be do them simultaneously, i.e. it's not specifically telling you which to do first.

So to start, I would (though I'm sure someone here has a much more clever trick, but I don't think it's necessary for these) start by writing out the general solution to

x+y-w=0

and x+y+2z-w=0

Because (keeping in mind we have two different subspaces), any vector of the subspace where s=0 must satisfy the first, and any vector of the subspace where s=2 satisfies the second

 

[radou]

Thanks Radou

You said ("Yes, you should" find the basis to find the dimension). But the question statement is as the following:" Find the dimension of the subspace and provide a basis for it", so here it is the other way around. I am really confused. Could you please clarify this point for me?

Well, first plug in the value of s for which the set is a subspace of R^4. Then, try to describe what a vector in the subspace must look like in order to satisfy the given equation (i.e. find the parametric solution of the equation).

 

[Neoon]

Lets choose x=1, y=1, w=2 in the case where s=0
x+y-w=1+1-2=0 which satisfy the equation
or
Choose x=2, y=2, w=4 in the case where s=0
x+y-w=1+1-2=0 which is also satisfy the equation
…etc

And choose x=1, y=1, z=1 w=4 in the case where s=2
x+y+sz-w=1+1+2(1)-4=0 which satisfy the equation
Or
Choose x=2, y=2, z=2 w=8 in the case where s=2
x+y+sz-w=1+1+2(1)-4=0 which is also satisfy the equation
…etc

So, we can generate many vectors. Which vectors to choose to be in the bases?

radou,

What is the parametric solution of the equation? I didn't understand

 

[Office_Shredder]

Neoon, the point is to find the most general solution.

For example, in the system a+b=0, b+c=0

we have a=-b
b=-c

So if we choose an arbitrary a, we know b=-a, and hence c=a, and we have completely solved the system (most general solution).

So x*(1,-1,-1) is a solution for any x.

So you need to find which variables can be arbitrary, and which ones are determined based on your choice.

 

[Neoon]

Office_Shredder,

I hope I understand what you want me to find.

In the case s=0
Say x = a, y = b lead to w = x + y = a + b,
w is determined based on x & y (a & b)

Or say x = a, w = b lead to y = w – x = b – a,
y is determined based on x & w (a & b)

Or finally say or say y = a, w = b lead to x = w – y = b – a,
x is determined based on y & w (a & b)


In the case s=2

Say x = a, y = b, z = c lead to w = x + y + 2z = a + b + 2c,
w is determined based on x, y & z (a, b & c)

Or say x = a, y = b, w = c lead to z = 1/2*(w-x-y) =1/2 (c-a-b),
z is determined based on x, y & w (a, b & c)

Or say x = a, z = b, w = c lead to y = (w-x-2z) =( c-a-2(b)),
y is determined based on x, z & w (a, b & c)

Or finally say y = a, z = b, w = c lead to x = (w-y-2z) =(c-a-2(b)),
x is determined based on y, z & w (a, b & c).


when are saying (x*(1,-1,-1) is a solution for any x.) what do mean by that ?

 

[Neoon]

I went and looked in other books and found a similar example.
In that example, the book found the rank first of a matrix containing all possible vectors.

So, I tried to find the dimension by first finding the rank as the following:

For s = 2:

Let x = a, y = b, z = c -- w = a+b+2c. So, we will have the following vector:
(a
B
C
A+b+2c)

Let x= a, y = b, w = c -- z = (1/2) * (c – a – b)
(a
B
(1/2)*(c-a-b)
c)

Let x= a, z = b, w = c -- y = c –a -2b
(a
c-a-2b
b
c)

Let y =a, z = b , w = c -- x = c – 2b – a
(c-2b-a
A
B
c)

Now, take a = 2, b = 3, c = 4 and we place the vectors as columns in a matrix:
(2 2 2 -4
3 3 -4 2
4 -1/2 3 3
13 4 4 4)

Reducing the above matrix gave me:
(2 2 2 -4
0 -9/2 -1 11
0 0 -7 8
0 0 0 0)

So, the rank is 3 and, hence, the dimension is 3.(there are only 3 indepent vectors)

I did the same thing for s=0 and i arrived to also 3 independent vectors.

Is my solution correct? please advise.

 

[Moo Of Doom]

As long as you're closed under addition and multiplication, you always have the zero vector in your set!

That's not quite true. Though the case where it fails can usually be quickly ruled out.

Can you guess why it is insufficient to just prove closure under linear combinations (closure under addition and scalar multiplication) to test for subspaces?

 

[Office_Shredder]

when are saying (x*(1,-1,-1) is a solution for any x.) what do mean by that ?

We decided that if (a,b,c) is a solution, then b=-a, c=a. So the solution really looks like (a,-a,a) for any a. We can factor out the a to get a*(1,-1,1) which is just the set of linear combinations of the basis of a straight line where the basis is (1,-1,1)

True moo, it needs to have a vector in it also

Neonn, you're on the right track in your first post.

Like you said, x+y=w solves the case of s=0. (note z can be whatever you want

So our solution looks like (x,y,z,x+y)

Or we can make that basis by separating it

x*(1,0,0,1) + y*(0,1,0,1) + z*(0,0,1,0) (you should see or be able to prove that's a basis) and you can see summming those vectors gives you a solution to the original equation for any x,y,z.

Do the same for s=2

 

[Neoon]

So the basis of the subspace when s=0 is:
x*(1,0,0,1), y*(0,1,0,1), z*(0,0,1,0)

And similarly the basis when s=2 is:
x*(1,0,0,1), y*(0,1,0,1), z*(0,0,1,2)

Both with dimension= 3

As per your comment before: I became an expert in subspaces, now, am I an expert in dimensions and bases or not yet?

 

[Office_Shredder]

That looks good to me

You can see how this method is much preferred to the one in your post (at least for simple problems like this).

 

[Neoon]

Thanks Office Shredder
Thanks very much. I really appreciate your valuable help