# Finding a matrix from a linear transformation

• Feb 5th 2013, 03:15 PM
jezb5
Finding a matrix from a linear transformation
not sure if this is classed as advanced algebra but here goes.

I need to find the matrix of t and I am confused as to which is the correct method to use.

The basis for the domain is E = {(1,0,0),(1,1,0),(0,1,1)} and the basis for the codomain is the standard basis of F = {(1,0,0),(0,1,0),(0,0,1)}

t : R^3 → R^3
(x,y,z) → (y - z, x + z, x + y)

This is what I have worked out so far:

I have found the images of the vectors in the domain with basis E = {(1,0,0),(1,1,0),(0,1,1)} to be

t(1,0,0) = (0,1,1)
t(1,1,0) = (1,1,2)
t(0,1,1) = (1,1,1)

as t(x,y,z) = (y - z, x + z, x + y)

and the F - coordinates of each of the image vectors, where F = {(1,0,0),(0,1,0),(0,0,1)}

t(1,0,0) = (0,1,1)F
t(1,1,0) = (1,1,2)F
t(0,1,1) = (1,1,1)F

so is the matrix of t with respect to the bases of E and F

0 1 1
1 1 2
1 1 1 (sorry, do not know how to do big brackets)

OR...........

do I need to use the below method for finding the matrix?

t(1,0,0) = (0,1,1) = (a,b,c)F
(a,b,c)F = a(1,0,0) + b(0,1,0), c(0,0,1) = (a,b,c)
a=1
b=0
c=0

t(1,1,0) = (1,1,2) = (d,e,f)F
(d,e,f)F = d(1,0,0) + e(0,1,0) + f(0,0,1) = (d,e,f)
d=1
e=1
f=0

t(0,1,1) = (1,1,1) = (g,h,i)F
(g,h,i)F = g(1,0,0) + h(0,1,0) + i(0,0,1) = (g,h,i)
g=0
h=1
i=1

and this gives the matrix of:

1 1 0
0 1 1
0 0 1

I think it is my first choice but need someone to clarify. I'm just getting to grips with the terminology so please be gentle with your answer http://www.mymathforum.com/images/smilies/icon_wink.gif
• Feb 5th 2013, 03:55 PM
chiro
Re: Finding a matrix from a linear transformation
Hey jezb5.

I used a change of basis formulation to get the answer and the answer I got was (with Octave):

>> E
E =

1 0 0
1 1 0
0 1 1

>> M
M =

0 1 -1
1 0 1
1 1 0

>> inv(E)*M
ans =e

0 1 -1
1 -1 2
0 2 -2

This assumes that the transformation is done in the basis of the domain and then a conversion to the standard basis is done afterwards.

I can walk you through the ideas if it is confusing.
• Feb 6th 2013, 01:55 AM
jezb5
Re: Finding a matrix from a linear transformation
If you could talk me through it, that'd be great, as it seems neither of my answers were correct. Could you see how I have done it, and explain that way? As I have these methods in a text book so it will be easier to refer back to. Thanks
• Feb 6th 2013, 11:09 AM
Deveno
Re: Finding a matrix from a linear transformation
the problem here is that there is more than one possible matrix for T. so we need to be clear on WHICH matrix for T we want.

the formula that defines T is probably "T in the standard basis" that is when you write T(x,y,z) = something, you mean (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1).

since our "something" in this case is (y-z,x+z,x+y) = (y-z)(1,0,0) + (x+z)(0,1,0) + (x+y)(0,0,1), using the standard basis F for both domain and co-domain we have:

$[T]_F^F = \begin{bmatrix}0&1&-1\\1&0&1\\1&1&0 \end{bmatrix}$

now if what we want is $[T]_F^E$, we can proceed as follows:

find the matrix P that changes E-coordinates to F-coordinates.

this matrix is easy to write down, its columns are simply the coordinates of the E-basis in the F-basis, so:

$P = \begin{bmatrix}1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$

so the matrix we seek is:

$[T]_F^E = [T]_F^FP = \begin{bmatrix}0&1&-1\\1&0&1\\1&1&0 \end{bmatrix} \begin{bmatrix}1&1&0\\0&1&1\\0&0&1 \end{bmatrix} = \begin{bmatrix}0&1&0\\1&1&1\\1&2&1 \end{bmatrix}$

ok, so: you now have 3 different answers to the same question, and you might wonder, how do i know which one is right?

well the first column of $[T]_F^E$ should be T(1,0,0), since (1,0,0) = [1,0,0]E.

and T(1,0,0) = (0-0,1+0,1+0) = (0,1,1).

and the second column of $[T]_F^E$ should be T(1,1,0), which is (1-0,1+0,1+1) = (1,1,2).

finally, the third column of $[T]_F^E$ should be T(0,1,1), which is (1-1,0+1,0+1) = (0,1,1).

**********

in more detail: we see that $[T]_F^E[v]_E$ = $[T]_F^E([a,b,c]_E) = [b,a+b+c,a+2b+c]_F$.

if we apply P-1 (that is express F-coordinates in terms of E-coordinates) to (x,y,z) = [x,y,z]F, we get [x,y,z]F = [x-y+z,y-z,z]E

(because [x,y,z]F = x[1,0,0]F + y[0,1,0]F + z[0,0,1]F, and:

(1,0,0) = [1,0,0]F = [1,0,0]E = 1(1,0,0) + 0(1,1,0) + 0(0,1,1)

(0,1,0) = [0,1,0]F = [-1,1,0]E = -1(1,0,0) + 1(1,1,0) + 0(0,1,1)

(0,0,1) = [0,0,1]F = [1,-1,1]E = 1(1,0,0) - 1(1,1,0) + 1(0,1,1), so that: [x,y,z]F = x[1,0,0]F + y[0,1,0]F + z[0,0,1]F

= x[1,0,0]E + y[-1,1,0]E = z[1,-1,1]E = [x,0,0]E + [-y,y,0]E + [z,-z,z]E = [x-y+z,y-z,z]E), so that:

$T(x,y,z) = [T]_F^E([x-y+z,y-z,z]_E)$

$= [y-z,(x-y+z)+(y-z)+z,(x-y+z)+2(y-z)+z]_F$

$= [y-z,x+z,x+y]_F = (y-z,x+z,x+y)$, as desired.
• Feb 6th 2013, 12:41 PM
jezb5
Re: Finding a matrix from a linear transformation
Well that is a much better explanation than my books. Thank you so much for that as I understand a lot better now (Happy)