Linear Algebra, Linear Transformations

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Unique linear transformations are defined by their action on a basis of the vector space, meaning that if the domain vectors form a basis, the transformation can be completely determined by a few values. In the examples provided, T(9x+5) and T(7x+4) are unique because they can be expressed in terms of independent polynomials, allowing for a unique mapping to vectors in R2. The transformation's linearity ensures that any linear combination of input vectors can be expressed as a linear combination of their corresponding outputs. The confusion arises from the interpretation of the transformations, where the notation and the nature of the polynomials involved need clarification. Understanding the basis and the linearity condition is key to grasping why certain transformations are unique.
auk411
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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.

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.
 
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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 R2 (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 R2 to R2.

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(au + bv) = 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.
 
Saying there is a unique linear tranformation satisfying certain conditions, means there is one and only one such transformation.

For example, if we are only told that the linear transformation L maps (1, 0), in R2 to (1, 1), then we know that L(x,y)= L(x(1, 0)+ y(0, 1))= xL(1,0)+ yL(0,1)= x(1,1)+ yL(0,1)= (x,x)+ yL(0,1). But I don't know what L(0,1) is. Since (0,1) and (1, 0) are independent, I cannot get L(0,1) from "L(1, 0)= (1,1)". There are many such (actually an infinite number) of Linear transformation satisfying that condition.

In general, a linear transformation on an n-dimensional vector space is completely determined by its action on n independent vectors (which then must be a basis for the space).

But if I am told that L maps (1, 0) to (1, 1) and maps (0, 1) to (2, 3), then I can say that L(x,y)= L(x(1,0)+ y(0,1))= xL(1,0)+ yL(0,1)=x(1,1)+ y(2,3)= (x+2y, x+ 3y). That is the only limear transformation satisfying those conditions- it is unique.

In the example, T(9x+5) = (.1,.2) and T(7x+4) = (.3,.8), "9x+ 5" and "7x+ 4" are independent and, since the set of linear polynomials has dimension 2, form a basis for that space. Given any such polynomial, ax+ b, it is easy to show that ax+ b= (4a- 7b)(9x+ 5)+ (9b-5a)(7x+ 4) so that T(ax+ b)= (4a- 7b)T(9x+5)+ (9b-5a)T(7x+4)= (4a+ 7b)(.1,.2)+ (9b-5a)(.3, .8)= (-1.1a+ 3.4b, -3.2a+ 8.6), a unique linear transformation.
 
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it is easy to show that ax+ b= (4a- 7b)(9x+ 5)+ (9b-5a)(7x+ 4) so that T(ax+ b)= (4a- 7b)T(9x+5)+ (9b-5a)T(7x+4)= (4a+ 7b)(.1,.2)+ (9b-5a)(.3, .8)= (-1.1a+ 3.4b, -3.2a+ 8.6), a unique linear transformation.


Basically, where you stop explaining and start handwaiving (see above) is where I need to more fully explain.

Please do show how to arrive at the conclusion you mention above, as well as where you get the values from (like 4a and 7b), along with why you set it up the way you did (say, why did you multiply (4a -7b)(9x+5).. etc?
 
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