Does This Set Form a Basis for R2?

[MienTommy]

Does this set span R_2? Is this set a basis for R_2? How many dimensions does it have?

1 -4
2 3
-4 6

This can be row reduced to
1 0
0 1
0 0

In row-reduced echelon form

I couldn't find the formatting to format this into a matrix, but these are column vectors.

 

[S.G. Janssens]

What do you think? Recall your definitions (of "span", "basis" and "dimension") and it should not be hard to answer this question. (BTW: Strictly speaking, the set of these two column vectors does not have any dimension, because the notion of "dimension" in an LA context applies only to linear (sub)spaces.)

 

[HallsofIvy]

I, personally, dislike setting vectors into matrices. I would prefer to use the basic definitions. Also, it can be confusing whether you intend the vectors to be the rows or columns. Actually, I think it is more common to do it as columns but, here, since you specify $R^2$, it is clear you intend the rows, (1, -4), (2, 3), and (-4, 6). One of the things you should know is that if the vector space has dimension "n", then NO set containing more than n vectors can be independent. (And no set containing fewer than n vectors can span the space.)

Here, you have three vectors while $R^2$ has dimension 2 so they cannot be independent and so cannot be a basis. The can still span the space. A set of vectors "spans" a space if and only if every vector in that space can be written as a linear combination of vectors in the spatce. Here, any vector in $R^2$ can be written as "(x, y)". Do there exist numbers, A, B, C, such that A(1, -4)+ B(2, 3)+ C(-4, 6)= (x, y). Can we find A, B, C such that A+ 2B- 4C= x and -4A+ 3B+ 6C= y for any x and y? If we multiply the first equation by 2 and add to the second (equivalent to your "row reduction") we cancel "A" and get 7B- 2C= 2x+ y. We now have only that single equation but that means that, given any value for B, we have C= (7/2)B- x- y/2. B can be any number, there are, in fact, an infinite number of such linear combinations (more than one is a consequence of this set NOT being independent). Yes, this set spans $R^2$. No, it is not a basis.[/B]

 

[S.G. Janssens]

it is clear you intend the rows, (1, -4), (2, 3), and (-4, 6).

I'm not sure that was the OP's intention, since

these are column vectors.

but well... this makes things even more obvious.

 

[MienTommy]

I, personally, dislike setting vectors into matrices. I would prefer to use the basic definitions. Also, it can be confusing whether you intend the vectors to be the rows or columns. Actually, I think it is more common to do it as columns but, here, since you specify $R^2$, it is clear you intend the rows, (1, -4), (2, 3), and (-4, 6). One of the things you should know is that if the vector space has dimension "n", then NO set containing more than n vectors can be independent. (And no set containing fewer than n vectors can span the space.)

Here, you have three vectors while $R^2$ has dimension 2 so they cannot be independent and so cannot be a basis. The can still span the space. A set of vectors "spans" a space if and only if every vector in that space can be written as a linear combination of vectors in the spatce. Here, any vector in $R^2$ can be written as "(x, y)". Do there exist numbers, A, B, C, such that A(1, -4)+ B(2, 3)+ C(-4, 6)= (x, y). Can we find A, B, C such that A+ 2B- 4C= x and -4A+ 3B+ 6C= y for any x and y? If we multiply the first equation by 2 and add to the second (equivalent to your "row reduction") we cancel "A" and get 7B- 2C= 2x+ y. We now have only that single equation but that means that, given any value for B, we have C= (7/2)B- x- y/2. B can be any number, there are, in fact, an infinite number of such linear combinations (more than one is a consequence of this set NOT being independent). Yes, this set spans $R^2$. No, it is not a basis.[/B]

I actually meant for them to be column vectors, so v_1 = [1 2 -4] v_2 = [-4 3 6]
I understand your explanation though. I just don't understand whether this is a basis or not if the vectors were column vectors. My guess is that it is, since it contains two pivots and is linearly independent, then it must be a basis for R_2. That last row is really throwing me off. I thought vectors in R_2 contain only two rows? Do the row sizes even matter?

 

[Mark44]

I actually meant for them to be column vectors, so v_1 = [1 2 -4] v_2 = [-4 3 6]
I understand your explanation though. I just don't understand whether this is a basis or not if the vectors were column vectors. My guess is that it is, since it contains two pivots and is linearly independent, then it must be a basis for R_2. That last row is really throwing me off. I thought vectors in R_2 contain only two rows? Do the row sizes even matter?

The vectors you show here are vectors in $\mathbb{R}^3$, so they definitely aren't in $\mathbb{R}^2$, so can't span it, and can't be a basis for it. Just by observation I can see that these two vectors are linearly independent, which means that they are a basis for and span a two-dimensional subspace of $\mathbb{R}^3$. Geometrically, this is a particular plane in three-dimensional space.

 

[MienTommy]

The vectors you show here are vectors in $\mathbb{R}^3$, so they definitely aren't in $\mathbb{R}^2$, so can't span it, and can't be a basis for it. Just by observation I can see that these two vectors are linearly independent, which means that they are a basis for and span a two-dimensional subspace of $\mathbb{R}^3$. Geometrically, this is a particular plane in three-dimensional space.

Ah, it seems the trouble I am having is knowing the size of the subspaces. So, an ℝ³ subspace is not in three dimensions? I've always thought ℝ² was a subspace in two dimensions, ℝ³ is a subspace in three dimensions, etc.. So the superscript k in ℝ^k only represents the rows in a column vector?

so if I have the following vectors: x = x₁ = [1 2 3 4 5 6]. x₂ = [2 2 1 3 4 5], x₃ = [1 2 9 9 2 1], x₄ = [5 9 3 2 1 6]
Written as column vectors [x₁ x₂ x₃ x₄] Could I say that the column space of A is in ℝ⁶? And the row space is in ℝ⁴? And it's null space is in ℝ⁴? And (if this set is linearly independent) its basis is in ℝ⁶?

I'm just confused by how the book writes the ℝ as ℝⁿ and ℝ^m. I know m represents the rows and n represents columns. In the Invertible Matrix Theorem, it states Col A = ℝⁿ for an n x n matrix. So is this referring to the rows or the columns? The other confusing part the was mentioned in a theorem was that a column space of an m x n matrix A is a subspace of ℝ^m.

Are column spaces of a matrix just its column vectors? Or is it the span of the column vector?

 

[Fredrik]

Ah, it seems the trouble I am having is knowing the size of the subspaces.

For a set S to be a spanning set for a vector space V, or a basis for V, it must be a subset of V. You asked if two vectors in $\mathbb R^3$ span $\mathbb R^2$. The set that contains those two vectors and nothing else, is a subspace of $\mathbb R^3$, not $\mathbb R^2$. Edit: I wrote that incorrectly. The sentence should say "spans a subspace", not "is a subspace".

So, an ℝ³ subspace is not in three dimensions?

$\mathbb R^3$ has exactly one 0-dimensional subspace, infinitely many 1-dimensional subspaces, infinitely many 2-dimensional subspaces, and exactly one 3-dimensional subspace. The 0-dimensional subspace is {0}. The 3-dimensional subspace is $\mathbb R^3$.

I've always thought ℝ² was a subspace in two dimensions, ℝ³ is a subspace in three dimensions, etc..

$\mathbb R^3$ is a 3-dimensional vector space. $\mathbb R^2$ is a 2-dimensional vector space. Every vector space is a subspace of itself.

So the superscript k in ℝ^k only represents the rows in a column vector?

A column vector has only one row. (Edit: Oops. I was pretty tired when I wrote that nonsense) The k tells you that the elements of the set are ordered n-tuples: $(x_1,\dots,x_n)$. They can be written either as $n\times 1$ matrices or as $1\times n$ matrices. The former option has a number of advantages that make it more popular.

I think it would be best if you study the basic definitions first (vector space, span, linearly independent, basis, subspace), and then ask your questions again.

 

[Mark44]

Ah, it seems the trouble I am having is knowing the size of the subspaces. So, an ℝ³ subspace is not in three dimensions?

There are many subspaces of the vector space $\mathbb{R}^3$. The subspace that consists only of <0, 0, 0> is zero-dimensional. Any line through the origin will be a one-dimensional subspace. Any plane through the origin will be a two-dimensional subspace. $\mathbb{R}^3$ is considered to be a subspace of itself. What is common to all of these subspaces is that any vectors in any of them have three coordinates.

I've always thought ℝ² was a subspace in two dimensions

$\mathbb{R}^2$ is a vector space. (It is also a subspace of itself.)

, ℝ³ is a subspace in three dimensions

$\mathbb{R}^2$ is a vector space. (It is also a subspace of itself.)

, etc.. So the superscript k in ℝ^k only represents the rows in a column vector?

Every vector in $\mathbb{R}^k$ will have k coordinates. Rows is not really the right term here.

so if I have the following vectors: x = x₁ = [1 2 3 4 5 6]. x₂ = [2 2 1 3 4 5], x₃ = [1 2 9 9 2 1], x₄ = [5 9 3 2 1 6]
Written as column vectors [x₁ x₂ x₃ x₄] Could I say that the column space of A is in ℝ⁶?

Yes, some subspace of $\mathbb{R}^6$. If they are linearly dependent, the dimension of the subspace they span will be less.

And the row space is in ℝ⁴?

Yes

And it's null space is in ℝ⁴?

Yes

And (if this set is linearly independent) its basis is in ℝ⁶?

Yes and no, assuming you're still talking about $x_1 \dots x_4$. When we talk about a basis, it's a basis for some subspace of whatever is the relavant vector space. We don't talk about the basis for a set of vectors.

I'm just confused by how the book writes the ℝ as ℝⁿ and ℝ^m. I know m represents the rows and n represents columns. In the Invertible Matrix Theorem, it states Col A = ℝⁿ for an n x n matrix.

This doesn't make sense to me.

So is this referring to the rows or the columns? The other confusing part the was mentioned in a theorem was that a column space of an m x n matrix A is a subspace of ℝ^m.

Are column spaces of a matrix just its column vectors?

The column space of a matrix is the subspace that is spanned by those vectors.

Or is it the span of the column vector?

 

[Mark44]

A column vector has only one row.

I disagree. A column vector has only on column. It could be thought of as an n x 1 matrix.

I think it would be best if you study the basic definitions first (vector space, span, linearly independent, basis, subspace), and then ask your questions again.

Good advice...

 

[HallsofIvy]

I'm not sure that was the OP's intention, since

but well... this makes things even more obvious.

I frankly didn't notice that. But the column vectors are in $R^3$ and the OP specifically talked at $R^2$.

 

[S.G. Janssens]

I frankly didn't notice that. But the column vectors are in ##R^3## and the OP specifically talked at ##R^2.##

Yes, I understand. When I wrote my initial reply I read the OP's question the other way around (i.e. I read that he was talking about $\mathbb{R}^3$ instead of $\mathbb{R}^2$). When I then read your reply, I realized that the answer to his question is even more obvious than I originally thought: the vectors are not even in the space.

 

[mfb]

Just a technical detail:

For a set S to be a spanning set for a vector space V, or a basis for V, it must be a subset of V. You asked if two vectors in $\mathbb R^3$ span $\mathbb R^2$. The set that contains those two vectors and nothing else, is a subspace of $\mathbb R^3$, not $\mathbb R^2$.

To get a subspace (instead of two isolated vectors), you have to take the span of that set of vectors.

 

[Fredrik]

I disagree. A column vector has only on column. It could be thought of as an n x 1 matrix.

Just a technical detail:To get a subspace (instead of two isolated vectors), you have to take the span of that set of vectors.

Thanks for catching those silly mistakes. This sort of thing happens a lot when I write a post a few hours past my bedtime.