You're welcome. Glad I could help.

[auk411]

Homework Statement

Will the values given below define a unique linear transformation? If so, find the value of T for an arb. domain element.

T(3,5) = (1,2) and T(2,3) = (6,7)

Homework Equations

The Attempt at a Solution

T(a[3,5]) + T(b[2,3]) = aT[3,5] + bT[2,3]
= a(1,2) + b(6,7)
=(a +6b, 2a +7)

This is wrong. But I have no idea how to even begin answering it. After reading the textbook, reading my notes, listening to the lecture audio, spending hours online reading, spending more hours just playing around with the problem, I am officially more lost and confused on the subject of linear transformations. I know what the answer is, btw. I just have no idea how to get there.

 

[Mark44]

Homework Statement

Will the values given below define a unique linear transformation? If so, find the value of T for an arb. domain element.

T(3,5) = (1,2) and T(2,3) = (6,7)

Homework Equations

The Attempt at a Solution

T(a[3,5]) + T(b[2,3]) = aT[3,5] + bT[2,3]
= a(1,2) + b(6,7)
=(a +6b, 2a +7)

This is wrong. But I have no idea how to even begin answering it. After reading the textbook, reading my notes, listening to the lecture audio, spending hours online reading, spending more hours just playing around with the problem, I am officially more lost and confused on the subject of linear transformations. I know what the answer is, btw. I just have no idea how to get there.

Here's a start.
1. Write (1, 0) as a linear combination of (3, 5) and (2, 3).
IOW, find constants a and b so that (1, 0) = a(3, 5) and + b(2, 3)
This entails solving the system
3a + 2b = 1
5a + 3b = 0

2. Write (0, 1) as a linear combination of (3, 5) and (2, 3).
IOW, find constants c and d so that (0,1) = c(3, 5) and + d(2, 3)
Similar to step 1.

 

[auk411]

Here's a start.
1. Write (1, 0) as a linear combination of (3, 5) and (2, 3).
IOW, find constants a and b so that (1, 0) = a(3, 5) and + b(2, 3)
This entails solving the system
3a + 2b = 1
5a + 3b = 0

2. Write (0, 1) as a linear combination of (3, 5) and (2, 3).
IOW, find constants c and d so that (0,1) = c(3, 5) and + d(2, 3)
Similar to step 1.

Thank you.

However, this really confuses me. WHY are we doing this? I have no examples in my book or by the teacher in class that do anything like this. So it is never mentioned that we should go about solving for (1,0) and (0,1). So even though I know your way leads to the right answer (I've worked out the problem your way) I have no idea WHY I am doing what I am doing.

 

[Mark44]

Because as I said and HallsOfIvy said in the other thread you started, if you know what a linear transformation does to its basis vectors, you know everything there is to know about the transformation. In particular, with this information, you can write a matrix that represents the linear transformation.

 

[auk411]

We already had a basis. So why do we need to find a new basis? Also, why are we trying to find a linear COMBINATION that equals, say, (1,0).

Couldn't we try to find a transition that takes in (2,3) and (3,5) and gives us (1,0)?

 

[Mark44]

We already had a basis. So why do we need to find a new basis? Also, why are we trying to find a linear COMBINATION that equals, say, (1,0).

Because some bases are easier to work with than others. In R², the standard basis is <1, 0>, <0, 1>. It's simple to decompose any vector in R² into a linear combination of these two vectors. For example, the vector <3, 2> can be written as 3<1, 0> + 2<0, 1>, and this can be obtained by inspection. How easy is it to write <3, 2> as a linear combination of <2, 3> and <3, 5>?

Couldn't we try to find a transition that takes in (2,3) and (3,5) and gives us (1,0)?

Essentially that's what we're doing, but that's a different transformation than the one you're trying to find.

 

[HallsofIvy]

We already had a basis. So why do we need to find a new basis? Also, why are we trying to find a linear COMBINATION that equals, say, (1,0).

Couldn't we try to find a transition that takes in (2,3) and (3,5) and gives us (1,0)?

Of course, we could! But that is not the problem you cited before. Also, note that since you have two different vectors being mapped to the same one, the linear transformation is not "one-to-one" and so is not invertible. In fact, it will map all of R² to the one dimensional subspace spanned by (1, 0): (x, 0) for x any number.

(1, 0)= a(2, 3)+ b(3, 5) gives 2a+ 3b= 1, 3a+ 5b= 0. Subtract twice the second equation from three times the first to eliminate a: -b= 3 so b= -3. Then 2a- 9= 1, a= 5. That is,
(1, 0)= 5(2,3)- 3(9,5) so that T(1, 0)= 5T(2, 3)- 3T(9, 5)= 5(1, 0)- 3(1, 0)= (2, 0). The first column of the matrix representing this linear transformation in the standard basis.

(0, 1)= a(2, 3)+ b(3, 9) gives 2a+ 3b= 0, 3a+ 5b= 1. Again, subtract twice the second equation from three times the first to eliminate a: -b= -2 so b= 2. Then 2a+ 6= 0 so a= -3.
(0, 1)= -3(2, 3)+ 2(3, 9) so that T(0, 1)= -3T(2, 3)+ 2T(3, 9)= -3(1, 0)+ 2(1, 0)= (-1, 0).

The matrix representing this linear transformation, in the standard basis, is
\begin{bmatrix}2 &amp; -1 \\ 0 &amp; 0\end{bmatrix}
Obviously that has 0 determinant and is not invertible. It is easy to see that the null space is given by (x, 2x) and the and the range by (x, 0) which is, as said before, spanned by (1, 0).

 

[auk411]

Because some bases are easier to work with than others. In R², the standard basis is <1, 0>, <0, 1>. It's simple to decompose any vector in R² into a linear combination of these two vectors. For example, the vector <3, 2> can be written as 3<1, 0> + 2<0, 1>, and this can be obtained by inspection. How easy is it to write <3, 2> as a linear combination of <2, 3> and <3, 5>?
Essentially that's what we're doing, but that's a different transformation than the one you're trying to find.

Thanks a billion times over!

 

[auk411]

Of course, we could! But that is not the problem you cited before. Also, note that since you have two different vectors being mapped to the same one, the linear transformation is not "one-to-one" and so is not invertible. In fact, it will map all of R² to the one dimensional subspace spanned by (1, 0): (x, 0) for x any number.

(1, 0)= a(2, 3)+ b(3, 5) gives 2a+ 3b= 1, 3a+ 5b= 0. Subtract twice the second equation from three times the first to eliminate a: -b= 3 so b= -3. Then 2a- 9= 1, a= 5. That is,
(1, 0)= 5(2,3)- 3(9,5) so that T(1, 0)= 5T(2, 3)- 3T(9, 5)= 5(1, 0)- 3(1, 0)= (2, 0). The first column of the matrix representing this linear transformation in the standard basis.

(0, 1)= a(2, 3)+ b(3, 9) gives 2a+ 3b= 0, 3a+ 5b= 1. Again, subtract twice the second equation from three times the first to eliminate a: -b= -2 so b= 2. Then 2a+ 6= 0 so a= -3.
(0, 1)= -3(2, 3)+ 2(3, 9) so that T(0, 1)= -3T(2, 3)+ 2T(3, 9)= -3(1, 0)+ 2(1, 0)= (-1, 0).

The matrix representing this linear transformation, in the standard basis, is
\begin{bmatrix}2 &amp; -1 \\ 0 &amp; 0\end{bmatrix}
Obviously that has 0 determinant and is not invertible. It is easy to see that the null space is given by (x, 2x) and the and the range by (x, 0) which is, as said before, spanned by (1, 0).

Thank you for many replies. I think you took my question the wrong way. But I still got something from your post.

thanks!