# Basis problem , finding coordinates

• Sep 15th 2012, 11:51 PM
Riazy
Basis problem , finding coordinates
Hi , the problem is written like this:
In a basis (e1,e2,e3) in r^3
F1=(1,0,1) , F2 = (0,1,1) f3 = (1,1,0)
Show That (F1,f2,f3) is also a basis in r^3 and give the relationship Between
the coordinates in both of the basis. Determine the coordinates for e1 + 2*f3

My progress:

Well first i use two fornulas

Y= t^-1 * x (1)

This is the formula to find coordinates of f basis
Where y is the basis f:s coordinates , and where t^-1
is the inverse of the change of basis matrix, and x is. The coordinates which belong
To the basis e

X= t*y. (2) T is the change of basis matrix

So

First of all i put up t

T is. 1 0 1
0 1 1
1 1 0

I have put the rows as columns

I find t through gauss elimination and get

T inverse. Is

1/2. *. 1 -1 1
-1 1 1
1 1 -1

The NeXT step is finding the coordinates

So i begin with formula. (1)

X = e1. And because. E2 and E3 is 0

Therefore the coordinates for x Will be
(1,0,0)

I use x And multiply it with t inverse
And get. (1/2 , -1/2 , 1/2). This is wrong according to the answer
It should be. (1/2, -1/2, 5/2) this in (e)

NeXT i go on using formula (2)

I multiply t with 2f3

And get. (2,2,0)

I add (2,2,0) + (1,0,0) = (3,2,0) this is correct

What am i doing wrong? Is there à better
Way to Solve this problen , pls aid me :)

Anywa
• Sep 16th 2012, 12:59 AM
Deveno
Re: Basis problem , finding coordinates
ok, we have two sets:

$B = \{e_1,e_2,e_3\} = \{(1,0,0),(0,1,0),(0,0,1)\}$ and

$B' = \{f_1,f_2,f_3\} = \{(1,0,1),(0,1,1),(1,1,0)\}$.

the first task is to show that B' is indeed a basis:

since we know $\mathrm{dim}_{\mathbb{R}}(\mathbb{R}^3) = 3$, it suffices to show that B' is linearly independent.

so suppose $c_1f_1 + c_2f_2 + c_3f_3 = (0,0,0)$, that is:

$c_1(1,0,1) + c_2(0,1,1) + c_3(1,1,0) = (c_1,0,c_1) + (0,c_2,c_2) + (c_3,c_3,0)$

$= (c_1+c_3,c_2+c_3,c_1+c_2) = (0,0,0)$. then:

$c_1+c_3 = 0$
$c_2+c_3 = 0$
$c_1+c_2 = 0$

hence $c_3 = -c_1 = -c_2$, so $c_1 = c_2$ thus $c_1+c_2 = 2c_1 = 0 \implies c_1 = 0$ and thus $c_2 = c_3 = 0$.

so B' is indeed linearly independent and thus a basis for $\mathbb{R}^3$.

now if T is the matrix that sends $[f_i]_{B'} \to [f_i]_B$ (so that $T([x]_{B'}) = [x]_B$),

then clearly the matrix that turns B-coordinates into B'-coordinates is $T^{-1}$.

you indeed have found T correctly, and so now we must find T-1.

we COULD use row-reduction to do this, but an easier way is to express the $e_i$ as linear combinations of the $f_i$.

now (1,0,0) = (1/2)(1,0,1) + (-1/2)(0,1,1) + (1/2)(1,1,0)

(0,1,0) = (-1/2)(1,0,1) + (1/2)(0,1,1) + (1/2)(1,1,0) and

(0,0,1) = (1/2)(1,0,1) + (1/2)(0,1,1) + (-1/2)(1,1,0), so:

$T^{-1} = \begin{bmatrix}\frac{1}{2}&-\frac{1}{2}&\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \end{bmatrix}$

this agrees with what you found.

now B-coordinates are just "the usual cooordinates" that is: $[(a,b,c)]_B = (a,b,c)$.

so the B-coordinates of $e_1 + 2f_3$ are: (1,0,0) + 2(1,1,0) = (1,0,0) + (2,2,0) = (3,2,0).

however, the B'-coordinates of $e_1 + 2f_3$ will be:

$T^{-1} \begin{bmatrix}3\\2\\0 \end{bmatrix}= \begin{bmatrix}\frac{3}{2}-\frac{2}{2}\\-\frac{3}{2}+\frac{1}{2}\\\frac{3}{2}+\frac{2}{2} \end{bmatrix} = \begin{bmatrix}\frac{1}{2}\\ -\frac{1}{2}\\ \frac{5}{2}\end{bmatrix}$,

so the answer you were given is correct (a sign error perhaps?)
• Sep 16th 2012, 01:17 AM
Riazy
Re: Basis problem , finding coordinates
Thank you man, the way you Solve basis and coordinates with, in which book can i read more about it, which book would u recommend, im using à swedish book