# Math Help - Scalars

1. ## Scalars

Having some LaTeX issues so this might look a bit off

let $e_1$= (1,0) and $e_2$ = (0,1) Show that for every $X$ $\in$ $R^2$, there exist unique scalars

$\alpha$ and $\beta$ such that $X$ = $\alpha$ $e_1$ + $\beta$ $e_2$

would this be $\alpha$(1,0) + $\beta$(0,1) = ( $\alpha$, 0) + (0, $\beta$) = ( $\alpha$, $\beta$) = $X$

I'll start there, there are two more parts to the question but I will see if I have this part first.

2. ## Re: Scalars

Originally Posted by Jonroberts74
Having some LaTeX issues so this might look a bit off

let $e_1$= (1,0) and $e_2$ = (0,1) Show that for every $X$ $\in$ $R^2$, there exist unique scalars

$\alpha$ and $\beta$ such that $X$ = $\alpha$ $e_1$ + $\beta$ $e_2$

would this be $\alpha$(1,0) + $\beta$(0,1) = ( $\alpha$, 0) + (0, $\beta$) = ( $\alpha$, $\beta$) = $X$

I'll start there, there are two more parts to the question but I will see if I have this part first.
ok there's a quick answer but this is important enough for a detailed answer

Your $e_1$ and $e_2$ form a basis for $R^2$

given any point in $R^2$ how do you compute what it's coordinates are with respect to this basis. You take the inner product of it and each of the basis vectors.

in this case <X, $e_1$> and <X, $e_2$> = $\alpha$ and $\beta$

then

X = $\alpha$ $e_1$+ $\beta$ $e_2$

3. ## Re: Scalars

okay the second part is

Let Y = (3, 0) and Z = (2, 0). Is it possible , for each vector X $\in$ $R^2$, to find $\alpha$ and $\beta$ such that X equals $\alpha$Y + $\beta$Z

4. ## Re: Scalars

Originally Posted by Jonroberts74
okay the second part is

Let Y = (3, 0) and Z = (2, 0). Is it possible , for each vector X $\in$ $R^2$, to find $\alpha$ and $\beta$ such that X equals $\alpha$Y + $\beta$Z
ok, you don't just get an answer for this.

why or why not would those two vectors serve as a basis for R2 ?

5. ## Re: Scalars

would it be No, because they are not linearly independent? they both run directly along the horizontal axis. Y = 2/3(z) They'd be in $R^1$

6. ## Re: Scalars

Yes. They aren't linearly independent so they cannot span R2 so the answer is no, you can't find alpha and beta for each vector in R2 such that X = alpha Y + beta Z

err no about the Y = 2/3 Z axis.

Both Y and Z are spanned by {1,0}. So any linear combination of Y and Z will still be spanned by {1,0}. These are the set of vertical vectors.

7. ## Re: Scalars

wait, I am confused now. My book is showing a cartesian plane but showing the first coordinate is the horizontal axis point and the second coordinate is the point of the vertical axis. In an illustration it is show x_1 and y_1 on the horizontal and x_2 and y_2 on the vertical. so by that, wouldn't Y then be equal to 2/3Z?

and the last part is

For Y = (1,2) and Z = (-4, -4) Find scalars alpha and beta such that X = alphaY +betaZ. Wouldn't they again just be scalar multiples of each other? so there would be none. Also does the fact that the second coordinate is merely the first doubled play any part into that?

8. ## Re: Scalars

the set of horizontal vectors I mean.

9. ## Re: Scalars

Originally Posted by Jonroberts74
wait, I am confused now. My book is showing a cartesian plane but showing the first coordinate is the horizontal axis point and the second coordinate is the point of the vertical axis. In an illustration it is show x_1 and y_1 on the horizontal and x_2 and y_2 on the vertical. so by that, wouldn't Y then be equal to 2/3Z?
Yes ok, Y = 3/2 Z but that only further shows you that the two vectors don't span R2. I guess I'm not sure what's confusing you.

10. ## Re: Scalars

sorry

meant $Y=\frac{3}{2}Z \Rightarrow (3,0) = \frac{3}{2}(2,0)$ was getting myself all mixed up

11. ## Re: Scalars

Originally Posted by Jonroberts74
and the last part is

For Y = (1,2) and Z = (-4, -4) Find scalars alpha and beta such that X = alphaY +betaZ. Wouldn't they again just be scalar multiples of each other? so there would be none. Also does the fact that the second coordinate is merely the first doubled play any part into that?
I'm a bit pressed for time at the moment so I'm going to refer you to a page that is on what this problem is about.

Change of Basis - HMC Calculus Tutorial

12. ## Re: Scalars

Originally Posted by romsek
I'm a bit pressed for time at the moment so I'm going to refer you to a page that is on what this problem is about.

Change of Basis - HMC Calculus Tutorial
what your problem is about is changing the basis from the column vectors e1={1,0} and e2={0, 1} to the column vectors v1={1, 2} and v2={-4, -4}.

I guess the first thing to do is ensure that we can use v1 and v2 as a basis. These two vectors must span R2 and thus be linearly independent.

The easiest way to verify this is to create a matrix of the column vectors v1 and v2 and find it's determinant. If it is non-zero then the two column vectors are linearly independent.

Here we get V = {{1,-4},{2,-4}} and det(V) = 4. Thus v1 and v2 are linearly independent and can be used as a basis for R2

Your original X was given in the first basis as X=a e1 + b e2

for the last part of your problem you want to express X in the new basis {1,2} {-4,-4} as some c v1 + d v2

You can write this all up using matrices of column vectors.

let E = {{1,0},{0,1}} i.e. made up of the column vectors e1 and e2. And let u = {a, b}

Then X = a e1 + b e2 = Eu

Similarly we can express the new basis as

V={{1,-4},{2,-4}}

and we'd like to find the components of w = {c, d} such that X = Vw, i.e. we want to express X in this new basis.

X = Vw = Eu

w = $V^{-1}Eu=V^{-1}u$

we can find the inverse of V easily enough

for a 2x2 matrix M={{m11, m12}, {m21, m22}} the inverse is given by 1/det(M){m22, -m12},{-m21,m11}}

so the inverse of V is 1/4{{-1,1},{-1/2,1/4}}

and our w = $V^{-1}u$ is given by

1/4{{-1,1},{-1/2,1/4}}{a,b} = {-a+b, -a/2+b/4} and so c = -a+b, d = -a/2 + b/4

In other words X =c v1 + d v2 as you were tasked to find.

You can verify this

(-a+b) v1 = {(-a+b),2(-a+b)} = {-a+b, -2a+2b }

(-a/2 + b/4) v2 = {-4(-a/2+b/4),-4(-a/2+b/4)} = {2a-b, 2a-b}

and summing we get

{-a+b, -2a+2b } + {2a-b, 2a-b} = {a, b} as expected (or hoped for anyway )

If your original basis wasn't E, and thus couldn't be just discarded in that matrix multiply to find w, then you'd just include it in that multiplication. You'd have it, it's just the column vectors of whatever basis you are going to originally describe X in.

See if any of this makes sense.

13. ## Re: Scalars

sorry meant (1,2) and (-4,-8). I was clearly not at my best last night.

also e_1 and e_2 are not related to (1,2) and (4,-8) in terms of the question. it was really more like three separate questions in one

14. ## Re: Scalars

well it's a quick write up on change of basis anyway. You can just apply everything in here to (1,2) (-4,8)