Thread: 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

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.

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

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:

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

now if what we want is $\displaystyle [T]_F^E$, 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:

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

so the matrix we seek is:

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

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 $\displaystyle [T]_F^E$ 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 $\displaystyle [T]_F^E$ should be T(1,1,0), which is (1-0,1+0,1+1) = (1,1,2).

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

**********

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

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:

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

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

$\displaystyle = [y-z,x+z,x+y]_F = (y-z,x+z,x+y)$, as desired.

Well that is a much better explanation than my books. Thank you so much for that as I understand a lot better now