# Describe Geometrically (line, plane, or all of R3) all linear combinations of:

#### cmm3594

##### New member
I know that someone posted this before, however I could not respond to that thread. I have not taken a Math course in several years and although I can do basic math and algebra, linear algebra is already seeming to be quite difficult. Basically with the title(subject line) the linear combinations are as follows:

(1, 2, 3) and (3, 6, 9) from what I read I understand that the second vector is just a multiple of 3 of the first vector and so the answer would be a line.

(1, 0, 0)and (0, 2, 3) although the back of the book has the answer I have no idea how to even approach this work. I'm not sure my professor is really explaining much of anything and expects everyone to know exactly what he is talking about.

(2, 0, 0) and (0, 2, 2) and (2, 2, 3)

If someone could please explain to me a good way to approach learning a new subject within mathematics I would greatly appreciate it.

V/R

Christina

#### caffeinemachine

##### Well-known member
MHB Math Scholar
I know that someone posted this before, however I could not respond to that thread. I have not taken a Math course in several years and although I can do basic math and algebra, linear algebra is already seeming to be quite difficult. Basically with the title(subject line) the linear combinations are as follows:

(1, 2, 3) and (3, 6, 9) from what I read I understand that the second vector is just a multiple of 3 of the first vector and so the answer would be a line.

(1, 0, 0)and (0, 2, 3) although the back of the book has the answer I have no idea how to even approach this work. I'm not sure my professor is really explaining much of anything and expects everyone to know exactly what he is talking about.

(2, 0, 0) and (0, 2, 2) and (2, 2, 3)

If someone could please explain to me a good way to approach learning a new subject within mathematics I would greatly appreciate it.

V/R

Christina
I am not sure what exactly is your question. Could you please phrase it again?

#### cmm3594

##### New member
I am not sure what exactly is your question. Could you please phrase it again?
Basically I'm asking how to geometrically identify what the vectors represent. i.e. how do i know that (1, 2, 3) and (3, 6, 9) geometrically represents a line.

#### Prove It

##### Well-known member
MHB Math Helper
Two points is enough to define a unique _____________?

Three points is enough to define a unique ______________?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Basically I'm asking how to geometrically identify what the vectors represent. i.e. how do i know that (1, 2, 3) and (3, 6, 9) geometrically represents a line.
The points $(1,2,3)$ and $(3,6,9)$ don't represent a line as such.
Although, It is true that a unique line passes through these two points.
Same is true for any two distinct points.
Does this help?

#### Deveno

##### Well-known member
MHB Math Scholar
It sounds like you don't really know exactly what a vector space is.

First and foremost, a vector space is some set. Usually this set is called $V$ if we need to give it a name, but don't want to be "too specific".

Secondly, we can "add vectors". This means we have some operation (denoted +), so that:

$u \in V, v \in V$ means that $u + v \in V$.

We require this operation to have the properties an addition ought to have:

V1. $u + (v + w) = (u + v) + w$ for all $u,v,w \in V$.
V2. $u + v = v + u$ for all $u,v \in V$
V3. There is some "0-vector" which satisfies:

$u + 0 = 0 + u = u$ for all $u \in V$.

V4. For each and every vector $u \in V$ there is another vector (usually written $-u$) with:

$u + -u = 0$.

For the set of triples $\Bbb R^3 = \{(x,y,z): x,y,z \in \Bbb R\}$ we have such an operation, given by:

$(x,y,z) + (x',y',z') = (x+x',y+y',z+z')$

and:

$0 = (0,0,0)$

with:

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

But that is not "all" that makes a vector space. In addition, we need a "field of scalars".

A field (typically called $F$, because mathematicians have no imagination) has its own addition, which satisfies similar rules:

F1. $a + (b + c) = (a + b) + c$ for all $a,b,c \in F$
F2. $a + b = b + a$ for all $a,b \in F$.
F3. There is an element $0 \in F$ which satisfies $a + 0 = 0 + a = a$ for all $a \in F$.
F4. For every element $a \in F$ there is some element (usually denoted $-a$) such that:

$a + -a = 0$.

Fields, however, also have a multiplication, which satisfies additional rules:

F5. $a(bc) = (ab)c$
F6. $ab = ba$
F7. There is an element $1 \in F$ with: $a1 = 1a = a$ for all $a$.
F8. For every $a \neq 0 \in F$, there is an element (usually denoted $1/a$) with:

$a(1/a) = 1$.

Finally, we have a "special way" addition and multiplication interact: the distributive law:

F9. $a(b + c) = ab + ac$.

Examples of fields include the rational numbers, the real numbers, and the complex numbers. It is important to understand that a vector space HAS to have the field specified, however, for most illustrations, the field of scalars is taken to be either the real numbers of the complex numbers, because these are well-understood (there are "bizarre" fields out there, just saying).

OK, so given our field $F$, we use scalars to "scale" vectors. Specifically the is a multiplication of the form:

scalar times vector = vector.

As you might suspect, this scalar multiplication also has to obey certain rules:

V5: $a(u + v) = au + av$ for all $a \in F, u,v \in V$.
V6: $(a + b)u = au + bu$ for all $a,b \in F, u \in V$. These are analogous to the distributive law in a field.

V7. $a(bu) = (ab)u$ for all $a,b \in F, u \in V$. This means that "successive scalings" can be calculated by taking the product of the scalars:

$u \to bu \to a(bu)$

gives the same result as:

$u \to (ab)u$.

Since by F6, $ab = ba$, we also have: $a(bu) = b(au)$.

V8. $1u = u$ for any $u \in V$.

Mixing up sums and scalar products gives what is called a linear combination:

$au + bv$ is a linear combination of $\{u,v\}$.

In $\Bbb R^3$ the scalar multiplication is given by:

$c(x,y,z) = (cx,cy,cz)$.

It is tedious, but straight-forward to verify $\Bbb R^3$ satisfies V1 through V8.

These rules (or "structure axioms") are absolutely necessary.

***********

Now, given a "parent" vector space, we are often interested in subspaces (subsets which are vector spaces in their own right). So which of V1 through V8 do we still have to verify so that a subset $U$ is also a vector space?

Well most of the properties will be "inherited" since if they hold for all vectors in $V$, they still hold for vectors in $U$ (since $U$ is IN $V$). In fact, we just have to check 3 things:

1. That vector addition restricted to $U$, gives an operation in $U$.

This is the same as saying:

1a. If $u,u' \in U$, then $u + u' \in U$ (closure of vector addition).

2. That scalar multiplication restricted to $U$, gives an operation on $U$.

This is the same as saying:

2a. If $u \in U, a \in F$, then $au \in U$ (closure of scalar multiplication).

Finally, since $U$ has to have a 0-vector (and $V$ only has one such), we require:

3. $0 \in U$.

***********

Often, we are just given a set of "basic vectors", and we want to construct a minimal subspace from that set. Such a subspace is said to be the SPAN of the given vectors.

What do we have to have for this to be a subspace?

Well, for any two vectors $v,v'$ in our original set, certainly $av$ and $bv'$ must be in this subspace (because of closure of scalar multiplication). And just as clearly, any sum of such scalar multiples must also be in the subspace, so $av + bv'$ must be in the subspace.

In fact, extending the same logic to any finite subset in our original set, we see that any linear combination of the vectors in our set must be in any subspace containing them. So, by our minimality requirement:

span(S) = all linear combinations of vectors in S.

***********

Given a set of vectors, S, we might ask: how many of them do we really need to generate the subspace they span?

Well suppose we have a set $S = \{v_1,v_2,v_3\}$ such that:

$v_3 = av_1 + bv_2$.

Clearly, we don't need $v_3$. We can re-write the above equation as:

$av_1 + bv_2 - v_3 = 0$.

In general, we say that if:

$c_1v_1 + c_2v_2 +\cdots+c_nv_n = 0$

with not all the $c_j = 0$ then $\{v_1,v_2,...,v_n\}$ is a LINEARLY DEPENDENT set.

In other words, one of the $v_j$ is a linear combination of the rest, so we have some "redundancy".

The "sister" concept is LINEAR INDEPENDENCE:

$c_1v_1 + c_2v_2 +\cdots+c_nv_n = 0 \implies c_1 = c_2 =\dots = c_n = 0$.

This expresses the fact that "we need every $v_j$".

I cannot under-state the importance of this concept, it is KEY to everything else you will do with vector spaces.

The maximum number of linearly independent vectors in a subspace of a vector space is called its DIMENSION. As you may well suspect, the dimension of $\Bbb R^3$ is 3.

A subspace of $\Bbb R^3$ will have dimension less than or equal to 3. You would do well to convince yourself why this is so.

The only 3-dimensional subspace of $\Bbb R^3$ is $\Bbb R^3$ itself.

A subspace of dimension 2 is called a PLANE.

A subspace of dimension 1 is called a LINE.

Note that not all lines or planes in $\Bbb R^3$ are subspaces in $\Bbb R^3$, they have to pass through the ORIGIN.

There is only ONE subspace of dimension 0 in $\Bbb R^3$, the single point, the origin.

*********

Now, given a set of $n$ vectors in $\Bbb R^3$ we can reduce it to a linearly independent subset of size 0,1,2 or 3, corresponding to the possible kinds of subspaces of $\Bbb R^3$. I will do this for a set of 4 vectors, so you can see how it is done.

Let $S = \{(1,0,1),(-2,1,1),(-1,1,2),(-2,2,4)\}$

Straight off the bat, it should be obvious that one of these vectors is unnecessary.

Indeed, (-2,2,4) = 2(-1,1,2), so clearly, we don't need (-2,2,4).

Let's see if we can find a,b such that:

(1,0,1) = a(-2,1,1) + b(-1,1,2) = (-2a-b,a+b,a+2b)

which is the same as the three equations:

-2a - b = 1
a + b = 0
a + 2b = 1

From the second equation, we have:

b = -a.

Substituting this in the other two equations:

-2a + a = 1
a - 2a = 1

both of which lead to a = -1. And indeed:

(1,0,1) = -(-2,1,1) + (-1,1,2).

So neither (1,0,1) nor (-2,2,4) is necessary, we can get them from linear combinations of the remaining two vectors.

As (-2,1,1) is not a scalar multiple of (-1,1,2), a maximal linearly independent subset has 2 elements, so the span is all linear combinations of:

(-2,1,1) and (-1,1,2), which forms a plane.

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I know that someone posted this before, however I could not respond to that thread. I have not taken a Math course in several years and although I can do basic math and algebra, linear algebra is already seeming to be quite difficult. Basically with the title(subject line) the linear combinations are as follows:

(1, 2, 3) and (3, 6, 9) from what I read I understand that the second vector is just a multiple of 3 of the first vector and so the answer would be a line.

(1, 0, 0)and (0, 2, 3) although the back of the book has the answer I have no idea how to even approach this work. I'm not sure my professor is really explaining much of anything and expects everyone to know exactly what he is talking about.

(2, 0, 0) and (0, 2, 2) and (2, 2, 3)

If someone could please explain to me a good way to approach learning a new subject within mathematics I would greatly appreciate it.

V/R

Christina
Hi Christina! Welcome to MHB!

If all vectors are a multiple of each other, they form a line through the origin.

If 2 vectors are independent, that is, not a multiple of each other, they "span" a plane.

If 3 vectors are independent, that is, the 3rd can not be written as the sum of multiples of the other 2 vectors, they "span" all of R3.
Note that if the 3rd vector can be written as the sum of multiples of the other 2 vectors, that vector effectively belongs to the plane those 2 vectors "span".
We say that the 3rd vector is a "linear combination" of the other 2.

#### Deveno

##### Well-known member
MHB Math Scholar
This page has a nice little picture of what the span of 2 vectors looks like:

Vector Span