Vector Subspace U+W = R^4: Solving with Homework Equations

[Physicsissuef]

Homework Statement

Are the vector subspaces U={(x,y,0,0) | x+2y=0} and W= {(0,0,z,t | z+t=0} from $$R^4$$ stand for U+W = $$R^4$$

Homework Equations

The Attempt at a Solution

Can somebody explain, how will solve this task. I have no idea, how they do in my book.
Thanks.

 

[HallsofIvy]

"Stand for"? Do you mean to ask "is the direct sum of U and W equal to R⁴"?

Each of U and V has dimension one so their direct sum will have dimension 2. It cannot possible by equal to R⁴!

 

[Physicsissuef]

"Stand for"? Do you mean to ask "is the direct sum of U and W equal to R⁴"?

Each of U and V has dimension one so their direct sum will have dimension 2. It cannot possible by equal to R⁴!

Why it cannot be possible by equal to R⁴? Can you explain how you solved, this task?

 

[quantumdude]

Think, man, think! What's the dimension of $\mathbb{R}^4$.

 

[Physicsissuef]

Think, man, think! What's the dimension of $\mathbb{R}^4$.

But aren't there any calculations. Sorry for my English, but I can't understand what you want to say.

 

[quantumdude]

OK, let's step through this. Do you understand why $dim(U \oplus W)=2$?

 

[Physicsissuef]

No, I don't understand it.

 

[quantumdude]

OK, let's write these subspaces another way. $U=span\{(-2,1,0,0)\}$ and $W=span\{(0,0,1,-1)\}$. Clearly, $dim(U)=dim(W)=1$. Since $U \cap W=\{(0,0,0,0)\}$, it follows that $U+W=span\{(-2,1,0,0),(0,0,1,-1)\}$. So, $dim(U+W)=2$.

What we actually have here is a direct sum $U \oplus W$ of the subspaces. If your texbook covers direct sums, then it probably also proves that the dimension of the direct sum of two subspaces equals the sum of the dimensions of the individual subspaces. But even if it doesn't, you can clearly see that $U+W$ has 2 independent vectors in its basis, so it has dimension 2.

So if you know what the dimension of $\mathbb{R}^4$ is, it's obvious why they can't possibly be equal.

 

[Physicsissuef]

OK, let's write these subspaces another way. $U=span\{(-2,1,0,0)\}$ and $W=span\{(0,0,1,-1)\}$. Clearly, $dim(U)=dim(W)=1$. Since $U \cap W=\{(0,0,0,0)\}$, it follows that $U+W=span\{(-2,1,0,0),(0,0,1,-1)\}$. So, $dim(U+W)=2$.

What we actually have here is a direct sum $U \oplus W$ of the subspaces. If your texbook covers direct sums, then it probably also proves that the dimension of the direct sum of two subspaces equals the sum of the dimensions of the individual subspaces. But even if it doesn't, you can clearly see that $U+W$ has 2 independent vectors in its basis, so it has dimension 2.

So if you know what the dimension of $\mathbb{R}^4$ is, it's obvious why they can't possibly be equal.

Ok, I understand know. Thank you very much. Just can you tell me what is dimension of $\mathbb{R}^4$?

 

[quantumdude]

If you don't know, I really think you should look it up.

 

[Physicsissuef]

If you don't know, I really think you should look it up.

Ok, dim$\mathbb{R}^4$=4

And if I have U and W vector subspaces.
U={ (x,y,z) | x+2y+3z=0 } and W= {(x,y,z) | x+y+z=0 }

Prove that $U \oplus W$ = $\mathbb{R}^3$

How will you solve this problem? Thanks.

 

[quantumdude]

Ok, dim$\mathbb{R}^4$=4

That's right. Now it should be clear why $U+W \neq \mathbb{R}^4$.

And if I have U and W vector subspaces.
U={ (x,y,z) | x+2y+3z=0 } and W= {(x,y,z) | x+y+z=0 }

Prove that $U \oplus W$ = $\mathbb{R}^3$

How will you solve this problem? Thanks.

Something is wrong with this problem. You've got two 2-dimensional subspaces of $\mathbb{R}^3$. Clearly, they cannot intersect in only the zero vector, so this isn't a direct sum. Perhaps you meant $U+W=\mathbb{R}^3$? If so, then I would begin by writing down the basis of each subspace and putting them into a list. This will give you a list of 4 vectors, and since they are elements of $\mathbb{R}^3$, at most 3 of them can be linearly independent. You need to find exactly 3 that are independent.

 

[Physicsissuef]

How you will find base for U or W?

 

[quantumdude]

That should be clear from the definitions of U and W. Each definition explicitly tells you what is required for a vector to belong to the subspace. Haven't you seen any examples of problems like this?

Here is a simple example.

Find a basis for the subspace $U=\{(x,y,z)|x=y\}$ of $\mathbb{R}^3$.

Solution:

An arbitrary vector $u\in U$ can be written as:

$$u=(x,x,z)$$

$$u=(x,x,0)+(0,0,z)$$

$$u=(1,1,0)x+(0,0,1)z$$

So clearly, $U=span\{(1,1,0),(0,0,1)\}$. Furthermore, $dim(U)=2$.

Get it?

 

[Physicsissuef]

Ok, I understand now. But for my case it is little complicated.

$U=\{(x,y,z)|x+2y+3z=0\}$ of $\mathbb{R}^3$.

$$u\in U$$

$$u=(x,\frac{x+2y-x}{2},\frac{-x-2y}{3})$$

$$u=(x,y,0)+(0,0,\frac{-x-2y}{3})$$

$$u=(x,0,0)+(0,y,0)+(0,0,\frac{-x-2y}{3})$$

$$u=(1,0,0)x+(0,1,0)y+(0,0,1)\frac{-x-2y}{3}$$

$U=span\{(1,0,0),(0,1,0),(0,0,1)\}$. So $dim(U)=3$

Now, for $$W$$.

$W=\{(x,y,z)|x+y+z=0\}$ of $\mathbb{R}^3$.

$$w\in W$$

$$w=(x,-x-z,-x-y)$$

$$w=(x,-x+x+y,-x-y)$$

$$w=(x,y,-x-y)$$

$$w=(x,y,0)+(0,0,-x-y)$$

$$w=(1,0,0)x+(0,1,0)y+(0,0,1)(-x-y)$$

$W=span\{(1,0,0),(0,1,0),(0,0,1)\}$. So $dim(W)=3$

$dim(W)+dim(U)=6$

Is this correct? Thanks.

 

[HallsofIvy]

No, it is not at all correct. How can you be expected to do problems like this and tell us that you do not know what "dimension" means and do not know what "direct sum" means? And, frankly, that is exactly what you are saying. U and W are both subsets of R³. Their direct sum is also a subset of R³ and so cannot have dimension higher than 3. Even if it were true that "dim(W)+ dim(U)= 6" but that tells you nothing about the dimension of their direct sum.

U= {(x,y,z)| x+ 2y+ 3z= 0}. You can, for example, solve that equation for x: x= -2y- 3z. Now, if we take y= 0, z= 1, we get x= -3 so <-3, 0, 1> is a vector in U. If we take y= 1, z= 0, we get x= -2 so <-2, 1, 0> is a vector in U. In fact it is easy to show that those two vectors form a basis for U (that's the advantage to using "0" and "1" for y and z) so the dimension of U is 2, not 3.

W= {(x,y,z)| x+ y+ z= 0}. z=-x-y so if we take x= 1, y= 0,we get z= -1. <1, 0, -1> is a vector in W. If we take x= 0,y= 1, we get z= -1 so <0, 1, -1> is a vector in W. Again, it is easy to see that they form a basis for W. W has dimension 2, not 3. But their direct sum is still a subspace of R³- the dimension of the subspace must be less than or equal to 3 so the sum, 2+ 2, is not the dimension of their direct sum.

The direct sum combines the bases. Since {<-2, 1, 0> , <-3, 0, 1>} is a basis for U and {<0, 1, -1>, <1, 0, -1>} is a basis for W, {<-2, 1, 0>, <-3, 0, 1>, <0, 1, -1>, <1, 0, -1>} spans their direct product. But that is not a basis because the four vectors are not independent. You need to find a subset of those 4 that are independent. The number of independent vectors that span the direct product is its dimension.

If you do not understand any of those words, like "independent" and "span", look them up in your textbook. They are essential to defining "dimension" of a vector space.

 

[Physicsissuef]

Ok, I understand what you want to say. Thanks. Let's say we have U and W subsets of$$R^3$$? Do always U and W must have dim=3? Can sometimes the dimension of U+W from $$R^3$$ be 4? Do must 3 of them to be linear combination of the 4-th of them? Or the 1st+2nd vector can be linear combination of the 4th one?

 

[Physicsissuef]

I have one new task.

Find $$U+W$$ $$\subseteq$$ $$\mathbb{R}^3$$

$$U=\{(a,b,0)| a=b,a,b$$ $$\in$$ $$\mathbb{R}\}$$

$$W=\{(0,b,c)| b=c, b,c$$ $$\in$$ $$\mathbb{R}\}$$

Can somebody help?

 

[Physicsissuef]

That's right. Now it should be clear why $U+W \neq \mathbb{R}^4$.



Something is wrong with this problem. You've got two 2-dimensional subspaces of $\mathbb{R}^3$. Clearly, they cannot intersect in only the zero vector, so this isn't a direct sum. Perhaps you meant $U+W=\mathbb{R}^3$? If so, then I would begin by writing down the basis of each subspace and putting them into a list. This will give you a list of 4 vectors, and since they are elements of $\mathbb{R}^3$, at most 3 of them can be linearly independent. You need to find exactly 3 that are independent.

How do you know that it will give me a list of 4 vectors?

 

[HallsofIvy]

Ok, I understand what you want to say. Thanks. Let's say we have U and W subsets of$$R^3$$? Do always U and W must have dim=3? Can sometimes the dimension of U+W from $$R^3$$ be 4? Do must 3 of them to be linear combination of the 4-th of them? Or the 1st+2nd vector can be linear combination of the 4th one?

If U and W are only sets and not subspaces they don't necessarily have any "dimension" at all! No, it is not true that subspaces of R³ must have dimension 3. It is true that they cannot have dimension greater than 3. As I said before, any combination of subspaces of R³ is still a subspace of R³ and cannot have dimension larger than 3. "must 3 of them to be linear combination of the 4-th of them" is the wrong way around. Since there cannot be more than 3 independent vectors in dimension 3, one of them must be a linear combination of the other 3. (I'm going to assume that you have a little problem with the English, but really understand what a "linear combination" is.) Of course, it might be that two of the vectors can be written as a linear combination the other 2 (in which case we have dimension 2) or that all are just some multiple of one (dimension 1).

I have one new task.

Find $$U+W$$ $$\subseteq$$ $$\mathbb{R}^3$$

$$U=\{(a,b,0)| a=b,a,b$$ $$\in$$ $$\mathbb{R}\}$$

$$W=\{(0,b,c)| b=c, b,c$$ $$\in$$ $$\mathbb{R}\}$$

Can somebody help?

Because "a= b", U is just the set of vectors of the form (a, a, 0)= a(1, 1, 0). That is one dimensional- {(1, 1, 0)} is a basis. Because "b= c", W is just the set of vectors of the form (0, b, b)= b(0, 1, 1). That is also one dimensional with basis {(0, 1, 1)}. Clearly those two vectors are independent- the only way we could have a(1, 1, 0)+ b(0, 1, 1)= (a, a, 0)+ (0, b, b)= (a, a+ b, b)= (0, 0, 0) is to have a= b= 0- so the two {(1, 1, 0), (0, 1, 1)} form a basis for U+ W which has dimension 2.

How do you know that it will give me a list of 4 vectors?

For each subspace you had one linear equation in x, y, z which you could solve for one in terms of the other two (say, z as a function of x and y). That means that by taking different values for x and y (I prefer x= 0, y= 1 and then x= 1, y= 0) we get two values of z, giving a basis of two vectors for each subspace. Those are the "4 vectors" Tom Mattson was referring to. Of course, they might not be independent so the set of all 4 might not form a basis for the direct sum.

 

[Physicsissuef]

Thanks. I have something that is unclear with the linear combination. If I have 4 vectors.
Lets say:
(1,2,0,0), (0,1,1,1), (2,4,0,0), (0,0,0,1)
Can I make:
2(1,2,0,0)=(2,4,0,0)
So the base will be (1,2,0,0),(0,1,1,1),(0,0,0,1) ?
And if W and U are sets, how can they have dimension?
Why we take two times different values for x and y for each subspace?
Thanks again for the help, and the efforts to help me.

 

[Physicsissuef]

HallsofIvy?

 

[HallsofIvy]

HallsofIvy?

I don't sit around waiting for questions! (My girlfriend just got back from overseas. I'm wondering why I'm sitting at a computer now!)

Thanks. I have something that is unclear with the linear combination. If I have 4 vectors.
Lets say:
(1,2,0,0), (0,1,1,1), (2,4,0,0), (0,0,0,1)
Can I make:
2(1,2,0,0)=(2,4,0,0)
So the base will be (1,2,0,0),(0,1,1,1),(0,0,0,1) ?

Yes, that is correct. Since you can get any thing involving (2, 4, 0, 0) using (1, 2, 0, 0) instead, you don't need it. Notice that you could just as easily use (2, 4, 0, 0) instead of (1, 2, 0, 0). The point is that you don't need both and a "basis" is the smallest set that will span your space.

And if W and U are sets, how can they have dimension?

Well, YOU were the one who referred to them as sets! And I said that they don't (necessarily) have a dimension. A "subspace" is a set of vectors that has special properties (closed under vector addition and scalar multiplication). Any subspace is a "set", not all sets are subspaces. Only vector spaces and their subspaces have "dimension".

Why we take two times different values for x and y for each subspace?
Thanks again for the help, and the efforts to help me.

My point was that we could. "x", "y" and any other such symbols are "dummies"- they just represent specific values and it doesn't matter if we call them "x", "y" or "a", "b" or anything else- as long as we are consistent. My point was that the vector space defined by {(x, y, 0)|x= y} is precisely the subspace of R³ in which the third component is 0 and the first two are the same. If you said {(a, b, 0}|a= b} or {(p, q, 0}|p= q} or even {(z,y,0)|z= y} you would mean exactly the same thing.

 

[Physicsissuef]

Ok, sorry.
I was asking about this part of your post:

For each subspace you had one linear equation in x, y, z which you could solve for one in terms of the other two (say, z as a function of x and y). That means that by taking different values for x and y (I prefer x= 0, y= 1 and then x= 1, y= 0) we get two values of z, giving a basis of two vectors for each subspace. Those are the "4 vectors" Tom Mattson was referring to. Of course, they might not be independent so the set of all 4 might not form a basis for the direct sum.

Why we get different values for x and y, 2 times (for the 1-st and then for the 2-nd subspace)? Why we don't get 3 times?

 

[Vid]

Basic Theorem on Dimension for Sums
Let U and W be subspaces of V
$$dim(U+W) = dim(U) + dim(W) - dim(U\cap W)$$

Since U and W are two planes that intersect on a line.

$$dim(U+W) = 2 + 2 - 1 = 3$$

 

[Physicsissuef]

No, it is not at all correct. How can you be expected to do problems like this and tell us that you do not know what "dimension" means and do not know what "direct sum" means? And, frankly, that is exactly what you are saying. U and W are both subsets of R³. Their direct sum is also a subset of R³ and so cannot have dimension higher than 3. Even if it were true that "dim(W)+ dim(U)= 6" but that tells you nothing about the dimension of their direct sum.

U= {(x,y,z)| x+ 2y+ 3z= 0}. You can, for example, solve that equation for x: x= -2y- 3z. Now, if we take y= 0, z= 1, we get x= -3 so <-3, 0, 1> is a vector in U. If we take y= 1, z= 0, we get x= -2 so <-2, 1, 0> is a vector in U. In fact it is easy to show that those two vectors form a basis for U (that's the advantage to using "0" and "1" for y and z) so the dimension of U is 2, not 3.

W= {(x,y,z)| x+ y+ z= 0}. z=-x-y so if we take x= 1, y= 0,we get z= -1. <1, 0, -1> is a vector in W. If we take x= 0,y= 1, we get z= -1 so <0, 1, -1> is a vector in W. Again, it is easy to see that they form a basis for W. W has dimension 2, not 3. But their direct sum is still a subspace of R³- the dimension of the subspace must be less than or equal to 3 so the sum, 2+ 2, is not the dimension of their direct sum.

The direct sum combines the bases. Since {<-2, 1, 0> , <-3, 0, 1>} is a basis for U and {<0, 1, -1>, <1, 0, -1>} is a basis for W, {<-2, 1, 0>, <-3, 0, 1>, <0, 1, -1>, <1, 0, -1>} spans their direct product. But that is not a basis because the four vectors are not independent. You need to find a subset of those 4 that are independent. The number of independent vectors that span the direct product is its dimension.

If you do not understand any of those words, like "independent" and "span", look them up in your textbook. They are essential to defining "dimension" of a vector space.

No, no, I am talking about this task. Why we get first
y=0, z=1 ; y=1, z=0(for the first subspace, that are 2 values for y and z)
x=1 y=0 ; x=0 y=1 (for the 2-nd subspace, that are 2 values for x and y)

Why don't we get 3 values for y and z or for x and y?

 

[Physicsissuef]

Can we get values, as much as we want ? How can I recognize how many times I should get how many values?

 

[HallsofIvy]

It should be relatively easy to see what the dimension is. R³, of course, has dimension 3- you can choose anyone of the three components to be whatever you want. If we have one condition, such as x- 3y+ z= 0, we can choose any value we want for two of the variables, then solve for the value of the third. That means the space is two dimensional. For example, z= 3y- x so choosing what ever values we want for x and y and z is fixed. If I chose, for example, x= y= 1, I would get z= 3- 1= 2. Obviously, <1, 1, 2> is a vector in that subspace. If I chose x= y= 2, I would get z= 6- 2= 4 so <2, 2, 4> is a vector in that subspace. But it is also clear that those vectors are not independent. One is just twice the other. By choosing x= 1, y= 0 and then x= 0, y= 1 I am certain to get independent vectors: <1, 0, -1> and <0, 1, 3>.

If I had a subspace of R₄ :<x, y, z, w> with 2x- 3y+ 4z- w= 0, we can solve for w= 2x- 3y+ 4z and see that this subspace is 3 dimensional. Choosing x= 1, y= z= 0 gives w= 2 so <1, 0, 0, 2> is a vector in that subspace. Choosing x= z= 0, y= 1, we get w= -3 so <0, 1, 0, -3> is a vector in that subspace. Finally, choosing x= y= 0, z= 1, we get w= 4 so <0, 0, 1, 4> is a vector in that subspace. And, in fact, those are a basis for the subspace. I get as many independent vectors that way as there are "free" variables. There is nothing "magic" about 0 and 1 except that they are simple and clearly not multiples of one another.

 

[Physicsissuef]

Ok, I have one more question. What if:
$U=\{(0,y,z)|2y+3z=0\}$ of $\mathbb{R}^3$
$W=\{(x,y,z)|x+y+z=0\}$ of $\mathbb{R}^3$
How we will find $$U\cap W$$?
I mean, is it possible?