Basis, what it really means for R

[flyingpig]

Homework Statement

Whenever LA talks about ℝⁿ, do they mean just the n?

Ex. Let's say I have two vectors

\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}

Now it is tempting to say that the two vector is a basis for ℝ².

Now my professor tells me that it isn't a basis for ℝ² because the vectors are in ℝ³. But I don't understand why. Sure they are in ℝ³, but the two vectors span ℝ².

Also, just to take another step further, let's say I have two vectors with three entries and I am telling you that they are linearly independent. My question is, is it a basis for ℝ²? Whenever we talk about ℝ², do we only mean the xy-plane and not some other plane that exist in ℝⁿ for n>2?

 

[Robert1986]

Any two linearly independent vectors span a 2-D subspace. As your prof said, the two vectors do not span R^2 because they are members of R^3. As you correctly pointed out, it is tempting to say that the two vectors span R^2. After all, the space spanned by them looks an awful lot like R^2. The subspace spanned by these two vectors isn't R^2, but it is isomorphic to R^2. That is, there is a 1-1, onto function from R^2 to the space spanned by your two vectors.

 

[lavinia]

Homework Statement

Whenever LA talks about ℝⁿ, do they mean just the n?

Ex. Let's say I have two vectors

\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}

Now it is tempting to say that the two vector is a basis for ℝ².

Now my professor tells me that it isn't a basis for ℝ² because the vectors are in ℝ³. But I don't understand why. Sure they are in ℝ³, but the two vectors span ℝ².

Also, just to take another step further, let's say I have two vectors with three entries and I am telling you that they are linearly independent. My question is, is it a basis for ℝ²? Whenever we talk about ℝ², do we only mean the xy-plane and not some other plane that exist in ℝⁿ for n>2?

Two independent vectors certainly span a 2 dimensional plane which is isomorphic to R^2. But R^2 is a special 2 dimensional plane that is defined as all pairs of real numbers.

 

[statdad]

"Now my professor tells me that it isn't a basis for ℝ2 because the vectors are in ℝ3. But I don't understand why. Sure they are in ℝ3, but the two vectors span ℝ2."

Bluntly put, these vectors are not in \mathbb{R}^2, and they don't span \mathbb{R}^2, because they are 3-dimensional vectors, not 2-dimensional vectors.

 

[HallsofIvy]

In particular, R^2 is NOT a subspace of R^3 and you seem to think it is. Yes, the set {(x, y, 0)} is a subspace of R^3 which is isomorphic to R^2 (like any two dimensional subspace) but it is NOT the same as R^2- (x, y) is NOT (x, y, 0).

 

[flyingpig]

I haven't learned the word "isomorphic". I think I can sort of understand a little about being members of ℝ³, but aren't they still on the xy plane? That shouldn't stop it from spanning the entire xy plane right?

 

[Robert1986]

But wait, you are now asking a different question. The two vectors most certainly span the xy-plane. However, the xy-plane in R^2 is not the xy-plane in R^3. When two sets are isomorphic, we mean that they are "essentially" the same, or that the structure is the same, or something similar (of course there is an actual definition, but this is just the intuition.) Thus, the xy-plane (indeed, every plane) in R^3 is essentially the same as the entire R^2 space. However, it is not the same because, if for no other reason, the vectors in R^3 are linear combinations of three basis vectors whereas those in R^2 are linear combinations of 2 basis vectors. Yes, one of the vectors happens to be the zero vector, but we still write the column vector with three entries.

 

[flyingpig]

But wait, you are now asking a different question. The two vectors most certainly span the xy-plane. However, the xy-plane in R^2 is not the xy-plane in R^3.

What do you mean they aren't the same. It's the same plane just drawn differently.

When two sets are isomorphic, we mean that they are "essentially" the same, or that the structure is the same, or something similar (of course there is an actual definition, but this is just the intuition.) Thus, the xy-plane (indeed, every plane) in R^3 is essentially the same as the entire R^2 space. However, it is not the same because, if for no other reason, the vectors in R^3 are linear combinations of three basis vectors whereas those in R^2 are linear combinations of 2 basis vectors. Yes, one of the vectors happens to be the zero vector, but we still write the column vector with three entries.

How does that justify it from being different though? It's same the plane!

 

[micromass]

I don't like myself saying this, but you're thinking much to intuitive. Normally that's a good thing, but now it prevents you to understand something.

\mathbb{R}^2 is defined as

\mathbb{R}^2=\{(x,y)~\vert~x,y\in\mathbb{R}\}

Intuitively, this is just a plane, but rigourously it is nothing more then a set of couples of reals.

On the other hand

\mathbb{R}^3=\{(x,y,z)~\vert~x,y,z\in \mathbb{R}\}

Intuitively again, this is a space. So you could argue that \mathbb{R}^2 is a plane contained in this space. But rigourously, this does not hold, since a vector (x,y) has two coordinates and thus cannot be in \mathbb{R}^3, which are vectors with 3 coordinates.

So, rigourously, \mathbb{R}^2 is not a subset of \mathbb{R}^3. However, intuitively, you would want it to be a subset. This is exactly why mathematicians have invented the word isomorphism. With that concept you can say that the two are related...

 

[flyingpig]

I don't like myself saying this, but you're thinking much to intuitive. Normally that's a good thing, but now it prevents you to understand something.

\mathbb{R}^2 is defined as

\mathbb{R}^2=\{(x,y)~\vert~x,y\in\mathbb{R}\}

Intuitively, this is just a plane, but rigourously it is nothing more then a set of couples of reals.

The xy-plane

On the other hand

\mathbb{R}^3=\{(x,y,z)~\vert~x,y,z\in \mathbb{R}\}

Intuitively again, this is a space.

The xyz-space

So you could argue that \mathbb{R}^2 is a plane contained in this space.

But it is! I can even draw it out for you! Here

http://img859.imageshack.us/i/10537124.png/

Uploaded with ImageShack.us

But rigourously, this does not hold, since a vector (x,y) has two coordinates and thus cannot be in \mathbb{R}^3, which are vectors with 3 coordinates.

Can't we just imagine or even know that the number 0 is in the last entry? I feel like this has more to deal with the concept of 0

So, rigourously, \mathbb{R}^2 is not a subset of \mathbb{R}^3. However, intuitively, you would want it to be a subset. This is exactly why mathematicians have invented the word isomorphism. With that concept you can say that the two are related...

I am trying to pronounce that word...iso-morpha-sem

 

[micromass]

Can't we just imagine or even know that the number 0 is in the last entry? I feel like this has more to deal with the concept of 0

No, this has nothing to do with the concept of 0! A number with two coordinates cannot be written as a number with three coordinates! A number (x,y) is not the same as (x,y,0). It's just an inconvenience of notation, but it's something we need to deal with...

If you want to see \mathbb{R}^2 as a subset of \mathbb{R}^3, then you'll have to use isomorphisms, there's no way around that!

 

[flyingpig]

I googled it and that word and the math that came with it was impossible for me...

When do I learn that word?

 

[micromass]

Probably right after you learn about linear functions. I don't know how long it will take your class to get there, but you will learn that word eventually

 

[flyingpig]

Still within Linear Algebra right?

 

[micromass]

Yes, linear functions between vector spaces is what linear algebra is all about! I have no doubt that you will learn about this in your linear algebra class!