# Two Matrix question about Span{ }

## Homework Statement

Question 1. Let $$v_{1} = \begin{bmatrix} 1\\ 0\\ -1\\ 0\\ \end{bmatrix}$$,$$v_{2} = \begin{bmatrix}0\\-1\\0\\1\\\end{bmatrix}$$, $$v_{3}=\begin{bmatrix} 1\\ 0\\ 0\\ -1\\ \end{bmatrix}$$

Does {v1, v2, v3} span $$\mathbb{R}^4$$ Why or why not?

Attempt at Question 1

Span is just a linear combination of all the vectors. So I simply just add up the vectors and get $$\begin{bmatrix} 2\\ -1\\ -1\\ 0\\ \end{bmatrix}$$

Since the last term is a 0, therefore this must stay in $$\mathbb{R}^3$$

Solution to Question one

[PLAIN]http://img87.imageshack.us/img87/5067/81744055.png [Broken]

Why are they instead asking if v3 is in the span of {v1, v2, v3}? What am I doing wrong?

Question 2

Let $$v_{1} = \begin{bmatrix} 0\\ 0\\ -2\\ \end{bmatrix}$$, $$v_{2}=\begin{bmatrix}0\\ -3\\ 8\\\end{bmatrix}$$, $$v_{3}=\begin{bmatrix}4\\ -1\\ -5\\ \end{bmatrix}$$

Does {v1, v2, v3} span $$\mathbb{R}^3$$ Why or why not?

Attempt

Notice that there is a {0,0,4} which means there is no solution.

If I were to use my original method and I add the vectors I get <4, -4, 1>

Which has three nonzero entries and therefore it spans in $$\mathbb{R}^3$$

My book is confusing me with the KEY

Solution by book

[PLAIN]http://img214.imageshack.us/img214/6066/81350323.png [Broken]

Last edited by a moderator:

Span is just a linear combination of all the vectors. So I simply just add up the vectors and get...

Your understanding of the definition of span is incorrect. The span of a set of vectors is the vector space generated by these vectors. An element of Span(v1, v2, v3) is a linear combination of these vectors, i.e. if v is in the span of these vectors, then there are real numbers $$a_1, a_2, a_3$$ such that $$v = a_1v_1 + a_2v_2 + a_3v_3$$. What the question is asking is if any vector in $$\mathbb{R}^4$$ can be written as a linear combination of the three vectors given. That is, given a vector v in $$\mathbb{R}^4$$, are there real numbers a1, a2, a3 such that $$v = a_1v_1 + a_2v_2 + a_3v_3$$? This will involve you solving a system of simultaneous equations, which is exactly what was shown in the answer key.

Also, I'd just like to note that there's a pretty egregious typo in the solution of the second problem. It should say that those vectors span $$\mathbb{R}^3$$, not $$\mathbb{R}^4$$.

Anyway, try looking over the definition of span again.

Your understanding of the definition of span is incorrect. The span of a set of vectors is the vector space generated by these vectors. An element of Span(v1, v2, v3) is a linear combination of these vectors, i.e. if v is in the span of these vectors, then there are real numbers $$a_1, a_2, a_3$$ such that $$v = a_1v_1 + a_2v_2 + a_3v_3$$. What the question is asking is if any vector in $$\mathbb{R}^4$$ can be written as a linear combination of the three vectors given. That is, given a vector v in $$\mathbb{R}^4$$, are there real numbers a1, a2, a3 such that $$v = a_1v_1 + a_2v_2 + a_3v_3$$? This will involve you solving a system of simultaneous equations, which is exactly what was shown in the answer key.

But there isn't even a v vector given

Last edited:
Mark44
Mentor
It should also be mentioned that a set of three vectors can't possibly span a four-dimensional space such as R4.
flyingpig said:
But there isn't even a v vector given
spamiam was just trying to explain what the term span meant. The span of a set of vectors v1, v2, v3 is the set of all possible linear combinations of those vectors. When you added the three vectors, you got the vector that was 1*v1 + 1*v2 + 1*v3. That's just one vector that's in the span of your three vectors.

It should also be mentioned that a set of three vectors can't possibly span a four-dimensional space such as R4.

Is it because 3 parameters?

spamiam was just trying to explain what the term span meant. The span of a set of vectors v1, v2, v3 is the set of all possible linear combinations of those vectors. When you added the three vectors, you got the vector that was 1*v1 + 1*v2 + 1*v3. That's just one vector that's in the span of your three vectors.

Yes, but the way they did it is like asking whether v3 is in the span of Span{v1, v2}

But there isn't even a v vector given

Exactly. You need to pick an arbitrary (i.e. unknown) vector v and see if it can be written as a linear combination. So set v = (x, y, z) for unknown x, y, and z, and then work out the system of simultaneous equations that you get from v = $$a_1 v_1 + a_2 v_2 + a_3 v_3$$ for arbitrary real numbers a1, a2, a3.

Is it because 3 parameters?

Yes, but the way they did it is like asking whether v3 is in the span of Span{v1, v2}

There's a result called the dimension theorem that can help you with these problems. I'll summarize it briefly, but you should probably take a look at the article below. Basically this theorem guarantees that if you have a set of n linearly independent vectors, then their span is an n-dimensional vector space.

Also, if you have three vectors, but one of them lies in the span of the other two (that is, is linearly dependent on the other two) than you can in effect "throw out" this vector since this vector is redundant: any linear combination involving all 3 vectors can also be written with just 2 of them. This means that the dimension of the vector space spanned by these three vectors is less than or equal to 2. And as you point out, $$\mathbb{R}^3$$ is 3-dimensional.

http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Anyway, try setting up equations for the second question. In fact with either method, you can't escape a system of simultaneous equations! It should also be mentioned that a set of three vectors can't possibly span a four-dimensional space such as R4.

Yeah, that's why the typo in the solution of the second problem is so glaring. But I'm not sure flyingpig has seen the dimension theorem.

Exactly. You need to pick an arbitrary (i.e. unknown) vector v and see if it can be written as a linear combination. So set v = (x, y, z) for unknown x, y, and z, and then work out the system of simultaneous equations that you get from v = $$a_1 v_1 + a_2 v_2 + a_3 v_3$$ for arbitrary real numbers a1, a2, a3.

I am still confused, shouldn't there be four entries (x,y,z,t)? Even so, that would give me EIGHT unknowns.

There's a result called the dimension theorem that can help you with these problems. I'll summarize it briefly, but you should probably take a look at the article below. Basically this theorem guarantees that if you have a set of n linearly independent vectors, then their span is an n-dimensional vector space.

Also, if you have three vectors, but one of them lies in the span of the other two (that is, is linearly dependent on the other two) than you can in effect "throw out" this vector since this vector is redundant: any linear combination involving all 3 vectors can also be written with just 2 of them. This means that the dimension of the vector space spanned by these three vectors is less than or equal to 2. And as you point out, $$\mathbb{R}^3$$ is 3-dimensional.

http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Anyway, try setting up equations for the second question. In fact with either method, you can't escape a system of simultaneous equations! Yeah, that's why the typo in the solution of the second problem is so glaring. But I'm not sure flyingpig has seen the dimension theorem.

I am sorry but that link just exploded my brain

Mark44
Mentor
Yes, the arbitrary vector should have four components.

No, there aren't eight variables - just three.
$$v = a_1 v_1 + a_2 v_2 + a_3 v_3$$
spamiam is assuming that you are given the three vectors v1, v2, and v3, as in your first post.

I am still confused, shouldn't there be four entries (x,y,z,t)? Even so, that would give me EIGHT unknowns.

Whoops! Sorry, you're right: you should have four entries. And as Mark44 said, you're given v1, v2, v3 in the problem.

I am sorry but that link just exploded my brain

I just meant that you should read the first paragraph. Basically this theorem says that any two bases of a vector space must have the same cardinality (i.e. number of elements). Since (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) form a basis for $$\mathbb{R}^4$$ that contains 4 vectors, then any basis for $$\mathbb{R}^4$$ must also have 4 vectors. Therefore no 3 vectors can generate $$\mathbb{R}^4$$.

Whoops! Sorry, you're right: you should have four entries. And as Mark44 said, you're given v1, v2, v3 in the problem.

I just meant that you should read the first paragraph. Basically this theorem says that any two bases of a vector space must have the same cardinality (i.e. number of elements). Since (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) form a basis for $$\mathbb{R}^4$$ that contains 4 vectors, then any basis for $$\mathbb{R}^4$$ must also have 4 vectors. Therefore no 3 vectors can generate $$\mathbb{R}^4$$.

basis...?

Mark44
Mentor
A basis for a vector space or subspace is a minimal spanning set. IOW, it is the smallest set of vectors that spans that space.

A basis is a linearly independent set that spans the vector space.

Here's a slightly more detailed definition from Wikipedia:

A basis B of a vector space V over $$\mathbb{R}$$ is a linearly independent subset of V that spans (or generates) V.

In more detail, suppose that $$B = \{ v_1, \ldots, v_n \}$$ is a finite subset of a vector space V over $$\mathbb{R}$$. Then B is a basis if it satisfies the following conditions:

* the linear independence property,

for all $$a_1, \ldots, a_n \in \mathbb{R}$$, if $$a_1v_1 + \cdots + a_nv_n = 0$$, then necessarily $$a_1 = \cdots = a_n = 0$$; and

* the spanning property,

for every x in V it is possible to choose $$a_1, \ldots, a_n \in \mathbb{R}$$ such that $$x = a_1v_1 + \cdots + a_nv_n$$.

This is turning to be even more difficult than the original question now lol because I jsut did Linear Dependence, TODAY lol

The original method I suggested will work and doesn't require any of the extra definitions or theorems in this post. Give it a try.