auk411 said:
Homework Statement
My question doesn't require numerical calculation. It is more about explanation.
Here it is: what does it mean to say there are unique linear transformations?
My textbook says "unique linear transformations can be defined by a few values, if the given domain vectors form a basis." However, that is all it says.
To "explain" here are 2 examples it gives: T(9x+5) = (.1,.2) and T(7x+4) = (.3,.8). This is a unique linear b/c (4a-7b)(9x+5) + (9b-5a)(7x+4) will equal ... [do some foiling] ... = ax +b for all a,b real. (I'm assuming that what the author is saying is that T(a,b) = (4a-7b)(9x+5) + (9b-5a)(7x+4) = ax +b for all a,b real.
[me: here's how lost I am. I'm not even sure why this is called a linear transformation and not linear transformationS. It looks like we are taking one transformation T_1 and doing something to it with T_2. Shouldn't we be asking why T(9x+5) is a unique linear transformation.
I've never seen linear transformations described as
unique linear transformations. In my experience, it's not something that usually comes up.
T(9x + 5) is not a linear transformation (unique or otherwise). T is the transformation, and in this problem T transforms a first degree polynomial to a vector in R
2 (i.e., a vector in the plane).
Since it requires two numbers to specify a first-degree polynomial, you could think of this transformation as one that transforms a pair of numbers (the coefficients of the polynomial) to another pair of numbers. In other words, you could think of this as a transformation from R
2 to R
2.
The polynomials could be represented as vectors such as <9, 5>. In this representation, the first coordinate represents the coefficient of x, and the second coordinate represents the constant. So T(<9, 5>) = <.1, .2> (Note that the vectors here should be considered to be column vectors.)
The vectors that you're given aren't very useful for characterizing this transformation. More useful would be <1, 0> and <0, 1>, or in terms of polynomials, x and 1. More about this later.
The thing about a linear transformation is its linearity; that is, if
u and
v are vectors in the domain of the tranformation, and a and b are scalars, then T(a
u + b
v) = aT(
u) + bT(
v).
We know that T(9x + 5) =T(<9, 5>) = <.1, .2>, and we know that T(7x + 4) = T(<7, 4>) = <.3, .8>
Then for a linear transformation, T(-4(9x + 5) + 5(7x + 4)) =T(-x) = -T(x) on the one hand, while T(-4(9x + 5) + 5(7x + 4)) = -4T(9x + 5) + 5T(7x + 4) = -4<.1, .2> + 5<.3, .8> = <1.1, 3.2>. From this I see that T(x) = <-1.1, -3.2>
Using something similar it's possible to find T(1). If I know T(x) and T(1) I know what T does to every 1st degree polynomial.
auk411 said:
Second example: T(2,1) = 4x +5; T(6,3) = 12x +15. Not unique because T(a, b) = 2ax + 5b works, but so does T(a,b) = 4bx + (3a - b).
I assume a and b are arbitrary domain elements. (please correct any assumptions if they are wrong). How does one arrive at the the formulas T(a, b) = 2ax + 5b and T(a,b) = 4bx + (3a - b)? What are these formulas telling me? What about the equation "T(a,b) = (4a-7b)(9x+5) + (9b-5a)(7x+4) = ax +b for all a,b real" tells me that this is a unique linear transformation? Could I just plug in anything to the a and b?
I'm lost. So anything will be helpful.