• Support PF! Buy your school textbooks, materials and every day products Here!

Linear Algebra- Subspaces

  • Thread starter Roni1985
  • Start date
  • #1
201
0

Homework Statement


Determine whether the following sets form subspaces of R^2 :

a) {(x1,x2)T | x1*x2=0}

b) {(x1,x2)T | x12=x22}

c) {(x1,x2)T | |x1|=|x2| }


Homework Equations





The Attempt at a Solution



My problem here is that I don't think I understand how the vectors look.
for instance,
{(x1,x2)T | x1*x2=0}

I say that each vectors looks like this:
[c,0]T
why zero ? because x2=0/x1
and x1=c.
Now, if this is correct, it must be a subspace. because we can multiply by a constant and still be in the subspace, or add two vectors and be in the subspace. But, its not a subspace according to the answers.

The problem with the other two is that I don't even know how the vectors in the subspaces look.

Would appreciate any help.
Thanks,
Roni.
 

Answers and Replies

  • #2
127
0
Yes. (ab=0) <=> ((a=0) or (b=0)) is true.

Note the "or".
Can you think of another general form of vectors in this set?
Then try add them (this isn't an uncommon course of action, ie, to guess two things which add and end up outside your set).
 
  • #3
201
0
Yes. (ab=0) <=> ((a=0) or (b=0)) is true.

Note the "or".
Can you think of another general form of vectors in this set?
Then try add them (this isn't an uncommon course of action, ie, to guess two things which add and end up outside your set).
I don't think I understand you.
How do you set the vectors ? this is the part that I don't understand.
I know how to test whether it forms a subspace or not once I have the general form of the vectors in the subspace.
 
  • #4
201
0
still need help :\

thanks.
 
  • #5
201
0
Yes. (ab=0) <=> ((a=0) or (b=0)) is true.

Note the "or".
Can you think of another general form of vectors in this set?
Then try add them (this isn't an uncommon course of action, ie, to guess two things which add and end up outside your set).
After reading your post again, I understand that I have to do something with the 'OR' but I don't really know how to set a matrix with OR .

I just came across another similar question:

{(x1,x2,x3)T | x3=x1 OR x3=x2 }

Don't really know what to do here.
 
  • #6
33,153
4,838
After reading your post again, I understand that I have to do something with the 'OR' but I don't really know how to set a matrix with OR .

I just came across another similar question:

{(x1,x2,x3)T | x3=x1 OR x3=x2 }

Don't really know what to do here.
Neither problem has anything to do with matrices. What's more, your question about "set(ting) a matrix with OR" doesn't make any sense.

The set in the first problem describes vectors in R2.
{(x1,x2)T | x1*x2=0}

The point the other poster was making was that if x1 * x2 = 0, then x1 = 0 or x2 = 0 (which includes the possibililty that both are zero). What do these vectors look like in R2? If you add any two vectors in this set, do you get another vector in this set? If you multiply any vector in this set by a scalar, do you get a vector in this set?

For the second problem, the set consists of vectors in R3. To get you started with this one, see if you can find some vectors in this set.
 
  • #7
201
0
Neither problem has anything to do with matrices. What's more, your question about "set(ting) a matrix with OR" doesn't make any sense.

The set in the first problem describes vectors in R2.
{(x1,x2)T | x1*x2=0}

The point the other poster was making was that if x1 * x2 = 0, then x1 = 0 or x2 = 0 (which includes the possibililty that both are zero). What do these vectors look like in R2? If you add any two vectors in this set, do you get another vector in this set? If you multiply any vector in this set by a scalar, do you get a vector in this set?

For the second problem, the set consists of vectors in R3. To get you started with this one, see if you can find some vectors in this set.
YEA I MEANT VECTORS, SORRY...
I HAVE THESE TYPES OF VECTORS [0,B]^T 0OR [C,0]^T
NOW IF I ADD ANY VECTOR OF TYPE [0,B]^T +[C,O]^T, I GET A VECTOR OUTSIDE OF THE SUBSPACE, I GET [C,B]^T....
THE THING IS THAT I WAS TRYING TO ADD [B,0]+[C,O]... AND IT DIDN'T WORK..

THANKS FOR THE HELP...

NOW HOW DO THE VECTORS LOOK HERE:
b) {(x1,x2)T | x12=x22}

HERE THEY MUST BE SQUARED ?
SAY
[C2,D2]^T ?

THANKS FOR YOUR HELP.
(SORRY FOR THE CAPS LOCK )
 
Last edited:
  • #8
33,153
4,838
YEA I MEANT VECTORS, SORRY...
I HAVE THESE TYPES OF VECTORS [0,B]^T 0OR [C,0]^T
NOW IF I ADD ANY VECTOR OF TYPE [0,B]^T +[C,O]^T, I GET A VECTOR OUTSIDE OF THE SUBSPACE, I GET [C,B]^T....
So what does that tell you about addition in this set?
THE THING IS THAT I WAS TRYING TO ADD [B,0]+[C,O]... AND IT DIDN'T WORK..
<B, 0> + <C, 0> = <B + C, 0>, which is a vector in this set.
THANKS FOR THE HELP...

NOW HOW DO THE VECTORS LOOK HERE:
b) {(x1,x2)T | x112=x122}
The set is {(x1,x2)T | x12=x22}
Can you find some specific vectors that belong to this set?
HERE THEY MUST BE SQUARED ?
SAY
[C2,D2]^T ?
Maybe, or maybe not. Look for some specific vectors in this set.
THANKS FOR YOUR HELP.
(SORRY FOR THE CAPS LOCK )
 
  • #9
201
0
so what does that tell you about addition in this set?
i got this one, it shows that it doesn't form a subspace, which is the correct answer

<b, 0> + <c, 0> = <b + c, 0>, which is a vector in this set.
right, i was doing only this type of vectors and was getting an answer that this is a subspace, which is incorrect.

the set is {(x1,x2)t | x12=x22}
can you find some specific vectors that belong to this set?
the thing that confuses me here is that, the vectors are of this form ? :
<c^2,d^2 > ?
We square both terms ? I think we do...

Say
<2^2,2^2> would be in the set ?
 
  • #10
33,153
4,838
mark44 said:
<b, 0> + <c, 0> = <b + c, 0>, which is a vector in this set.
right, i was doing only this type of vectors and was getting an answer that this is a subspace, which is incorrect.
The thing is, all vectors in the set have to add to another vector in the set. In your first example, <0, b> + <c, 0> don't add to a vector in the set, so even though some selected vectors add to a vector in the set, there are some that don't.
the thing that confuses me here is that, the vectors are of this form ? :
<c^2,d^2 > ?
We square both terms ? I think we do...

Say
<2^2,2^2> would be in the set ?
Yes, this vector is in the set. A vector <x1, x2> is in the set iff x12 = x12.

Here are some other vectors. Can you tell which of these are in the set?
<1, 2>
<2, 2>
<3, -3>
<9, 16>
<-3, -3>
<0, 0>

Keep in mind that you aren't to the point where you're deciding whether the set is a subspace. You're at the point where you're trying to find out which vectors are in the set.
 
  • #11
201
0
Oh I see...
then,
<2, 2>
<3, -3>
<-3, -3>
<0, 0>

are in the set....
but if we add 2 vectors
say
<3, -3>
<-3, -3>
we get a vector outside of the set.Therefore, it doesn't form a subspace.

Thanks very much for your help.
take care....
 

Related Threads for: Linear Algebra- Subspaces

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
2
Replies
31
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
Top