Thread: Help with deriving a formula for a linear transformation T

Hey guys, doing some linear transformation and I got stuck.
Here is the problem exactly as written in the book:

The action of a linear transformation T on a basis for R^3 is given. Derive a formula for T.

T([1, 0, 1]) = [0, 1]

T([0, -1, 1]) = [1, 0]

T([1, -1, 0]) = [0, 0]

This is all that is given.
For the problem before this, the transformation was from R^2 to R^2, and I managed to solve it by creating a system of equations and then solving for T(x1) and T(x2).
But the same method does not work here, since in this case its transforming from R^3 to R^2 and I'm not sure how to accommodate for that.

Thanks for the help.


P.S. - the answer is actually given in the back of the book, here it is for reference:


T([x1, x2, x3]) = [(-1/2)x1 - (1/2)x2 + (1/2)x3] [(1/2)x1 + (1/2)x2 + (1/2)x3]


the right side is first row and second row, just not sure how to format it on here yet.

Thanks again.

Okay, so we have a linear transformation T: R^3 --> R^2. We are then searching for a matrix A, since the transformation is defined by T(x) = Ax. Since we know that x is a column vector (3 by 1) and that the product of Ax is a 2 by 1 column vector we're searching for a matrix A which is a 2 by 3 matrix.

A =
[a b c]
[d e f]

We also have that

T([1;0;1]) = [0;1]
T([0;-1;1]) = [1;0]
T([1;-1;0]) = [0;0]

Using the fact that T(x) = Ax we can get a system of equation for the three given transformations. Multiply the yet unknown matrix A with the entries a,b,c,d,e,f. It will look like:

a*1+0*b+1*c = 0
d*1+e*0+1*f = 1
a*0+b*-1+c*1 = 1
d*0+e*-1+f*1 = 0
a*1+b*-1+c*0 = 0
d*1+e*-1+f*0 = 0

You can write this as an augmented matrix with 6 unknowns
[1 0 1 0 0 0 0]
[0 0 0 1 0 1 1]
[0 -1 1 0 0 0 1]
[0 0 0 0 -1 1 0]
[1 -1 0 0 0 0 0]
[0 0 0 1 -1 0 0]

Which is row equivalent to

[1 0 0 0 0 0 -1/2]
[0 1 0 0 0 0 -1/2]
[0 0 1 0 0 0 1/2]
[0 0 0 1 0 0 1/2]
[0 0 0 0 1 0 1/2]
[0 0 0 0 0 1 1/2]

We then have

a = -1/2
b = -1/2
c = 1/2
d = 1/2
e = 1/2
f = 1/2

A =
[-1/2 -1/2 1/2]
[ 1/2 1/2 1/2]

Multiply this with the columnvector [x1;x2;x3] and you have your given transformation

in this case, what we'd actually like, in order to get a formula, is to find what T does to the standard basis (because then we could just take a vector as a linear combination of the standard basis vectors, and use the SAME linear combination of T of those basis vectors).

unfortunately, we don't KNOW what T does to the standard basis vectors ( which are (1,0,0), (0,1,0) and (0,0,1), just for future reference).

but...(bright idea!)...suppose we expressed the standard basis vectors (which are, after all, still elements of R³) in the basis B = {(1,0,1),(0,-1,1),(1,-1,0)}.

so let's figure out how to DO just that.

what we are looking for, first, is a,b,c such that a(1,0,1) + b(0,-1,1) + c(1,-1,0) = (1,0,0). this is the same as:

(a+c,-b-c,a+b) = (1,0,0). that is:

a+c = 1
-b-c = 0
a+b = 0

well, the 3rd equation tells us we can replace b with -a. then our first two equations are:

a+c = 1
a-c = 0

add these together, and we get:

2a = 1, so a = 1/2. thus b = -1/2, and c = 1/2. let's check this:

(1/2)(1,0,1) + (-1/2)(0,-1,1) + (1/2)(1,-1,0) = (1/2,0,1/2) + (0,1/2,-1/2) + (1/2,-1/2,0) = (1,0,0) yippee!

now we do the same thing for (0,1,0):

a(1,0,1) + b(0,-1,1) + c(1,-1,0) = (0,1,0)

(a+c,-b-c,a+b) = (0,1,0)

a+c = 0
-b-c = 1
a+b = 0

again b = -a, and we have:

a+c = 0
a-c = 0

2a = 1, so a= 1/2. then b = -1/2, and c = -1/2. you can check that these coefficients work.

doing the same thing with (a+c,-b-c,a+b) = (0,0,1) gives us:

a+c = 0
-b-c = 0
a+b = 1

here, b = 1-a, so we get:

a+c = 0
(a-1)-c = 0 <=> a-c = 1.

so a = 1/2, b = 1/2, c = -1/2.

NOW, we can find a "standard" formula for T:

T(x,y,z) = T(x(1,0,0) + y(0,1,0) + z(0,0,1)) = T[(x/2)((1,0,1) - (0,-1,1) + (1,-1,0)) + (y/2)((1,0,1) - (0,-1,1) - (1,-1,0)) + (z/2)((1,0,1) + (0,-1,1) - (1,-1,0))]

= (x/2)T((1,0,1)) - (x/2)T((0,-1,1)) + (x/2)T((1,-1,0)) + (y/2)T((1,0,1)) - (y/2)T((0,-1,1)) + (y/2)T((1,-1,0)) + (z/2)T((1,0,1)) + (z/2)T((0,-1,1)) - (z/2)T((1,-1,0)) (since T is linear)

= ((x+y+z)/2)T((1,0,1)) + ((z-x-y)/2)T((0,-1,1)) + ((x+y-z)/2)T((1,-1,0)) (collecting like terms)

now we KNOW what T((1,0,1)), T((0,-1,1)) and T((1,-1,0)) are, they are (0,1), (1,0) and (0,0) respectively, so

= ((x+y+z)/2))(0,1) + ((z-x-y)/2))(1,0)

= ((z-x-y)/2,(x+y+z)/2), which agrees with the answer in your book (letting x₁ = x, x₂ = y, and x₃ = z).

Hey, thanks for the solution, it makes sense. The only part which I don't understand is why/how are you equating 1*a+0*b+1*c = 0 and again to 1 with different coefficients?

Originally Posted by fkf

[INDENT]a*1+0*b+1*c = 0
d*1+e*0+1*f = 1

Hey, thanks for the solution, it makes sense. The only part which I don't understand is why/how are you equating 1*a+0*b+1*c = 0 and again to 1 with different coefficients?

i was "re-using" the variables a,b,c since after we found them for one basis vector (1,0,0), i didn't see the point in making NEW variables for (0,1,0) and (0,0,1). perhaps i should have used a',b',c' and a",b",c" to avoid confusion.

I was actually asking about fkf's solution :P

As for yours though, I was able to follow how you were using the basis, until the part where you started finding the standard formula for T.

well suppose we have a space of dimension 2 (so it won't be as messy to write) and we have 2 bases A = {a₁,a₂} and B = {b₁,b₂}.

and suppose that we are given (for a linear transformation T) T(a₁) and T(a₂) but we WANT T(b₁) and T(b₂).

suppose further that we know:

b₁ = c₁a₁+c₂a₂
b₂ = d₁a₁+d₁a₂

then:

T(b₁) = T(c₁a₁+c₂a₂) = c₁T(a₁) + c₂T(a₂)
T(b₂) = T(d₁a₁+d₂a₂) = d₁T(a₁) + d₂T(a₂)

and since we're GIVEN the T(a₁), T(a₂) we can now CALCULATE the T(b₁), T(b₂).

so if our vectors "normally" come in a certain basis (the vector (x₁,x₂) is just an "abbreviation for" x₁e₁ + x₂e₂ = x₁(1,0) + x₂(0,1)) but we're given them in a second basis, we have to somehow "transform" one basis into another

(that's what the c's and d's do above).

so what i did is:

A) figure out what the standard basis vectors are in, in the "2nd basis" (which is {(1,0,1),(0,-1,1),(1,-1,0)}).

B) re-write (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) as x((1,0,0) in the 2nd basis) + y((0,1,0) in the 2nd basis) + z((0,1,0) in the 2nd basis)

that's 3 scalar multiples of linear combinations of 3 "2nd basis vectors" (9 terms in all).

C) take T of everything.

D) use the linearity of T to get 3 scalar multiples of linear combinations of 3 "T of 2nd basis vectors" (still 9 terms)

E) since we only have 3 different "T of some 2nd basis vector" combine the similar ones (put all the T(1,0,1) together, all the T(0,-1,1) together, all the T(1,-1,0) together).

F) now we only have 3 terms...but it turns out the all the T(1,-1,0) are 0...bye bye!...2 terms left:

something*T(1,-1,0) + (something else)*T(0,-1,1)

G) but T(1,-1,0) = (0,1) and T(0,-1,1) = (1,0) and:

something*(0,1) + (something else)*(1,0) = (something else)*(1,0) + something*(0,1) = (something else,something)

that is: (1,0) and (0,1) are the standard basis for R², so (u,v) = u(1,0) + v(0,1)

Originally Posted by zdravkoBG

Hey, thanks for the solution, it makes sense. The only part which I don't understand is why/how are you equating 1*a+0*b+1*c = 0 and again to 1 with different coefficients?

Because we have six unknowns. We need to have six (different) equations to solve the system of equations. Every line gives information about what a,b,c,d,e,f should be.