To recap: A vector is an ordered list of components, also referred to as an n-tuple, where n is the number of components. For example, (A, B, C) is a vector with three components, also called a 3-tuple.

Scalar multiplication multiplies each component of a vector with a scalar, the same scalar multiplying each component, resulting in a scaled vector:

a( f, g, h ) = ( af, ag, ah )

A linear combination is the sum of scaled vectors, each vector scaled with a scalar that may be different than the scalars used to scale the other vectors:

a₁𝐯₁ + a₂𝐯₂ + ⋯ + an𝐯n

And a vector space is a set of vectors that are linear combinations of each other.

An important characteristic of a vector space is closure: All vectors in a vector space are linear combinations of other vectors in the space; and all vectors that are linear combinations of vectors in the space remain in the space.

Formally, denoting a vector space as V, with 𝐮 and 𝐯 denoting vectors that are in the vector space V, the following two properties hold:

1.

2.

2.

𝐮 + 𝐯 ∈ V

λ𝐮 ∈ V

λ𝐮 ∈ V

where λ is a scalar, and ∈ symbolizes “is an element of”.

Property 1 states that the sum of two vectors is in the same vector space as the two vectors that are summed.

Property 2 states that a scaled vector is in the same vector space as the vector before it was scaled.

[+] Show more formal properties notation

Vectors that meet these two conditions form a vector space.

In previous lessons, we have been using the following general form of the equation of a line:

Ax + By = C

where the slope of the line is −(A/B), and the y-intercept is C/B.

We used the coefficients A, B and C of that line equation as vector components, in that order, to specify each line as a vector (A, B, C).

And we are finding that linear combinations of those vectors generate lines that have the same intersection, with those (A, B, C) vectors forming a vector space.

Consider the following lines from the previous lesson o:

Figure 1: Lines intersecting at a point (in this case the origin) have equations with coefficients that form a vector space.

The equations of these lines are:

𝑥 + 𝑦

−(½)𝑥 + 𝑦

(½)𝑥 + 2𝑦

2𝑥 − 𝑦

−(½)𝑥 + 𝑦

(½)𝑥 + 2𝑦

2𝑥 − 𝑦

= 0

= 0

= 0

= 0

= 0

= 0

= 0

and the coefficients A, B and C of these equations as vectors (A, B, C) are:

( 1, 1, 0 )

(-1/2, 1, 0 )

( 1/2, 2, 0 )

( 2, 1, 0 )

Since the y-intercept of this form of line equation is C/B, and the lines intersect at the origin, where y=0, the y-intercept C/B must then be zero, hence C must be zero, as shown in those vectors of coefficients.

Notice that there is no way to take linear combinations of these vectors, or any other (A, B, C) vector that has C=0, to produce a vector with C not equal to zero, proving that all linear combinations of these vectors (this vector space) will generate lines that intersect the origin.

This can be proven for intersection points other than the origin by translating (offsetting) the intersection point to the origin with change of variables (temporarily forming a “new origin”).

Change of variables, when possible, allows us to temporarily replace a variable with a more simple variable to help solve an equation.

For example, consider factoring the following polynomial:

x6 − 2x3y3 + y6

We recall the following formula from the Elementary Algebra lesson:

(x − y)2 = (x − y)(x − y) = x2 − 2xy + y2

At first glance, that formula may not appear useful in this case. But wait! There is a way to use that formula.

Using the following formula, also from the Elementary Algebra lesson:

(Xa)b = Xab

the polynomial we are trying to factor is equivalent to:

(x3)2 − 2x3y3 + (y3)2

This combination of exponents is suitable for us to use change of variables to substitute x for x3 and to substitue y for y3, yielding the following polynomial that is easier to factor:

x2 − 2xy + y2

Factoring using the above mention formula yields:

(x − y)2

Then substituting x3 back in for x and substituting y3 back in for y:

(x3 − y3)2

Using change of variables to temporarily translate (offset) an intersection point on a plane to the origin is possible because the translation does not change properties of the lines. The results can simply be translated back by the same offsets in the opposite directions.

We illustrate with an example. Consider the following lines from the Linear Equations lesson:

Figure 2: Lines intersecting at (2, 3)

The coefficients of the red line are the component-wise sum (vector addition) of the coefficients of the black lines.

The equations for those black lines are:

𝑥 + 𝑦

−(½)𝑥 + 𝑦

−(½)𝑥 + 𝑦

= 5

= 2

= 2

Translating both of those lines by substituting x + 2 and y + 3 for x and y respectively (which translates the x-axis and y-axis in the +2 and +3 directions to put the origin under the intersection point — essentially the same as moving the intersection point to the origin):

𝑥 + 𝑦

(𝑥+2) + (𝑦+3)

𝑥 + 2 + 𝑦 + 3

𝑥 + 𝑦 + 5

𝑥 + 𝑦

−(½)𝑥 + 𝑦

−(½)(𝑥+2) + (𝑦+3)

(−½)𝑥 + (−½)2 + 𝑦 + 3

(−½)𝑥 + (−1) + 𝑦 + 3

(−½)𝑥 − 1 + 𝑦 + 3

(−½)𝑥 + 𝑦 + 2

(−½)𝑥 + 𝑦

(𝑥+2) + (𝑦+3)

𝑥 + 2 + 𝑦 + 3

𝑥 + 𝑦 + 5

𝑥 + 𝑦

−(½)𝑥 + 𝑦

−(½)(𝑥+2) + (𝑦+3)

(−½)𝑥 + (−½)2 + 𝑦 + 3

(−½)𝑥 + (−1) + 𝑦 + 3

(−½)𝑥 − 1 + 𝑦 + 3

(−½)𝑥 + 𝑦 + 2

(−½)𝑥 + 𝑦

= 5

= 5

= 5

= 5

= 0

= 2

= 2

= 2

= 2

= 2

= 2

= 0

= 5

= 5

= 5

= 0

= 2

= 2

= 2

= 2

= 2

= 2

= 0

yields the line equations:

𝑥 + 𝑦

−(½)𝑥 + 𝑦

−(½)𝑥 + 𝑦

= 0

= 0

= 0

which are the black lines through the origin in Figure 1 above.

The red line through the origin in Figure 1, generated by component-wise addition of the coefficients of those black lines through the origin, has this equation:

(½)𝑥 + 2𝑦 = 0

Translating the coordinate frame back (changing the variables back) puts the intersection point at (2, 3) again, by substituting x−2 and y−3 for x and y in the equation of that red line (moving the coordinate axes −2 and −3 from under the red line):

(½)𝑥 + 2𝑦

(½)(𝑥−2) + 2(𝑦−3)

(½)𝑥 − (½)2 + 2𝑦 − 6

(½)𝑥 − 2/2 + 2𝑦 − 6

(½)𝑥 − 1 + 2𝑦 − 6

(½)𝑥 + 2𝑦 − 7

(½)𝑥 + 2𝑦

(½)(𝑥−2) + 2(𝑦−3)

(½)𝑥 − (½)2 + 2𝑦 − 6

(½)𝑥 − 2/2 + 2𝑦 − 6

(½)𝑥 − 1 + 2𝑦 − 6

(½)𝑥 + 2𝑦 − 7

(½)𝑥 + 2𝑦

= 0

= 0

= 0

= 0

= 0

= 0

= 7

= 0

= 0

= 0

= 0

= 0

= 7

which gives us the red line (½)x + 2y = 7 of Figure 2 above, showing the same result whether the linear combination created the red line at (2, 3) or at the origin. Indeed, the point of intersection could have been temporarily translated to any point.

Linear Equations

Vectors

Vector Space (this page)

Dimensions

Vectors

Vector Space (this page)

Dimensions

Copyright © 2020 Arc Math Software, All rights reserved

Arc Math Software, P.O. Box 221190, Sacramento CA 95822 USA Contact

2020–Aug–11 22:07 UTC

Arc Math Software, P.O. Box 221190, Sacramento CA 95822 USA Contact

2020–Aug–11 22:07 UTC