Linear Map f:R^2 to R^3 with Given Inputs (1,2) and (2,1)

[Physicsissuef]

Homework Statement

Find the linear map f:R^2 \rightarrow R^3, with f(1,2) = (2,1,0) and f(2,1)=(0,1,2)

Homework Equations

The Attempt at a Solution

I actually don't understand this task. PLease help! Thank you...

 

[eys_physics]

Hi
If you don't know where to begin, use that a linear map can be written:
\[\left(\begin{array}{c}f_{1} &amp; f_{2} &amp; f_{3}\end{array}\right)=<br /> F\left(\begin{array}{c}x_{1} &amp; x_{2}\end{array}\right)\]
Where F is a two-by-three matrix.

 

[HallsofIvy]

(1, 2) and (2, 1) are independent vectors and so form a basis for R². Knowing what a linear map does to a basis tells you what it does to any vector.

You need to show what f(x,y) is for any pair (x, y). To do that write (x, y) as a linear combination of (1, 2) and (2, 1): That is, find a, b so that a(1, 2)+ b(2, 1)= (x, y). Then f(x,y)= af(1,2)+ bf(2,1).

 

[Physicsissuef]

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

like this?

 

[HallsofIvy]

Does that look anything like "a(1, 2)+ b(2,1)= (x,y)"?

a(1, 2)+ b(2, 1)= (a, 2a)+ (2b, b)= (a+ 2b, 2a+ b)= (x, y) so a+ 2b= x, 2a+ b= y. Solve that for a and b in terms of x and y.

 

[Physicsissuef]

a=1
b=1
x=3
y=3
like this?

 

[HallsofIvy]

NO! Find a and b for any x and y!

 

[Physicsissuef]

x=1
y=1

a+2b=1

2a+b=1

should I make system out of those equations?

Actually, I don't understand the point of this task... I don't know what should I find...

 

[HallsofIvy]

Well, I've told you twice already: Solve the equations a+ 2b= x, 2a+ b= y for a and b. you should get a= an expression with x and y in it, b= an expression with x and y in it. Once you have done that you will be able to say that (x, y)= a(1, 2)+ b(2, 1) so that
f(x,y)= af(1,2)+ bf(2,1) and you know what f(1,2) and f(2,1) are.

You want to be able to find f(x,y) for any x and y. Do not assume x= y= 1!

 

[Physicsissuef]

a=-\frac{x-2y}{3}

b=\frac{2x-y}{3}
,
now what to do next?

 

[dynamicsolo]

We've dropped a dimension somewhere, haven't we? How does this map get the third coordinate?

The approach eys_physics suggested might be more straightforward. Write your argument vector and your resulting vector as either row or column vectors. If you choose row vectors, you have

[1 2] · [first row: a b c , second row: d e f] = [2 1 0] ,

and similarly for the other set. You will have a 2 x 1 matrix (2-d row vector) times a 2 x 3 mapping matrix giving you a 3 x 1 matrix (3-d row vector). Following the rules of matrix multiplication will give you a set of three equations in two distinct variables (the coefficients of the mapping matrix) for the first transformation shown, and a second set of three equations for the transformation of [2 1] to [0 1 2].

You now have a pair of equations relating a and d, another pair for b and e, and a third pair for c and f. All of these sets are simple to solve.

 

[Physicsissuef]

Actually, I don't know why you multiply by [1 2] and why it is equal to [2 1 0]

 

[HallsofIvy]

a=-\frac{x-2y}{3}

b=\frac{2x-y}{3}
,
now what to do next?

Excellent!

Actually, I don't know why you multiply by [1 2] and why it is equal to [2 1 0]

Haven't we already been through what a "basis" is? R² has dimension 2 and, since (1, 2) and (2, 1) are two independent vectors (one is not a multiple of the other) they form a basis. Every vector in can be written as a linear combination of those two vectors. You have now determined how they can be written: for any x, y,
(x,y)= \frac{2y-x}{3}(1, 2)+ \frac{2x-y}{3}(2, 1)
Since f is a linear map, by definition of "linear",
f(x,y)= \frac{2y-x}{3}f(1, 2)+ \frac{2x-y}{3}f(2, 1)
Now, you are told in the original problem that f(1,2)= (2, 1, 0) and that f(2, 1)= (0, 1, 2).

f(x,y)= \frac{2y-x}{3}(2, 1, 0)+ \frac{2x-y}{3}(0, 1, 2)
= \left(\frac{4y- 2x}{3},\frac{2y-x}{3}+ \frac{2x-y}{3},\frac{4x- 2y}{3}\right)
= \left(\frac{4y-2x}{3},\frac{x+y}{3},\frac{4x-2y}{3}\right)

 

[Physicsissuef]

Ok, I understand now how you find it. But I can't understand what we do actually to find it.
f(x,y)= af(1,2)+ bf(2,1)

f(x,y)= a(2,1,0)+ b(0,1,2)

we actually make linear combination of the 2 vectors in the basis, why?

 

[HallsofIvy]

Ok, I understand now how you find it. But I can't understand what we do actually to find it.
f(x,y)= af(1,2)+ bf(2,1)

f(x,y)= a(2,1,0)+ b(0,1,2)[/quote

we actually make linear combination of the 2 vectors in the basis, why?

Because they were the only ones ones for which we knew what f does!

I hope that, by this time, you understand what a basis is.

 

[Physicsissuef]

Is it (x,y)= a(2,1,0)+ b(0,1,2) or (x,y)= a(1,2)+ b(2,1)