# Geometric interpretation of an equation

1. Nov 23, 2011

### Vai

1. The problem statement, all variables and given/known data

x,y, z are vectors in R^n. We have the equation:

ax +by +cz, where a,b,c are constants such that a+b+c=1, and a,b,c>=0

What is the geometric interpretation of the equation?

2. Relevant equations

sv + tu, where u,v are vectors in R^n and s,t are constants such that s+t=1, s,t>=0 geometrically represents a line between u,v.

3. The attempt at a solution

So by similar logic, would ax +by +cz be a line through x,y,z?

2. Nov 24, 2011

### Ray Vickson

For (a): you do not have an "equation". An equation is something with an equal sign in it, and you don't have that. However, you do have an *expression*. The expression ax +by +cz is a convex combination of x, y and z, and so is a vector in the planar triangle formed by the three endpoints of x, y and z.

RGV

3. Nov 24, 2011

### Harrisonized

I'm guessing that you mean the vector Av, where bold denotes column vectors.

A=[x y z]
v=(a,b,c)

The product Av can be interpreted in two ways.

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The first way is to assume vector addition. A vector is an object in ℝn that denotes a coordinate in ℝn.

You can represent it by a point, or by an arrow that points from an origin to that point. The arrow is fictitious, but it shows how vector addition works. When vectors are added, you simply add each component in the vector to arrive at a new vector.

Three vectors (assuming they are linearly independent) in ℝn span a parallelepiped in ℝn. The origin is a coordinate on the parallelepiped, and each of the three vectors represent the coordinate of each adjacent corner of the parallelepiped. The sum x+y+z gives the coordinate of the far corner of the parallelepiped spanned by x, y, and z.

If we let v be a vector of weights, then the resulting product Av is still the coordinate of the far corner of a parallelepiped, but v distorts the side lengths of the parallelepiped spanned by x, y, and z.

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The second way is to view v, the vector of weights, as points in ℝ3. Then Av is a linear transformation on v that takes points out of ℝ3 and puts them in ℝn. Since v=(a,b,c) and a+b+c=1, the domain of v is a cube with corners ±e1, ±e2, ±e3, where e represents a standard unit vector. The corresponding range of Av in ℝn is a pseudocube in ℝn spanned by the bases x, y, z. Note that n linearly independent columns are necessary to span ℝn, so the range of this transformation is only a subset of ℝn.

Linear transformations preserve "structure". What that means is that a linear transformation can only distort, scale, or rotate, assuming the columns of A are linearly independent. It cannot make domains bend back on each other. In other words, the output of a linear transformation is always unique to its input, unless the columns of A are linearly dependent.

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Fyi, an equation has an equal sign. I thought this was common sense.

Last edited: Nov 24, 2011