One-to-One Linear Transformations

[_N3WTON_]

Homework Statement

Give a thorough explanation as to why a linear transformation:

with a standard matrix A CANNOT be one to one.

Homework Equations

The Attempt at a Solution

I think I have figured this one out, but I was hoping somebody could confirm whether this example is sufficient:

There are two equations with three unknowns. Therefore, this system cannot have a unique solution, instead it will have an infinite number of solutions. Therefore, there cannot be a one-to-one mapping from

 

[Stephen Tashi]

There are two equations with three unknowns. Therefore, this system cannot have a unique solution,

If you are using "a,b,c" to represent scalars, it doesn't make sense to put arrows over them.

It's possible for some systems of two equations in three unknowns to have a unique solution (or no solutions) so you would have to give more specifics to argue along those lines. You need to show explicitly that there are two different elements in \mathbb{R}^3 that are mapped to the same element in \mathbb{R}^2.

It's hard for me to guess what approach the problem expects you take, since I don't know what material has been covered prior to its being asked.

 

[_N3WTON_]

If you are using "a,b,c" to represent scalars, it doesn't make sense to put arrows over them.

It's possible for some systems of two equations in three unknowns to have a unique solution (or no solutions) so you would have to give more specifics to argue along those lines. You need to show explicitly that there are two different elements in \mathbb{R}^3 that are mapped to the same element in \mathbb{R}^2.

It's hard for me to guess what approach the problem expects you take, since I don't know what material has been covered prior to its being asked.

The instructor gave a hint that we are supposed to use pivots to prove the statement is true, but I'm not really seeing how to do that...

 

[Mark44]

Homework Statement

Give a thorough explanation as to why a linear transformation:

with a standard matrix A CANNOT be one to one.

Homework Equations

The Attempt at a Solution

I think I have figured this one out, but I was hoping somebody could confirm whether this example is sufficient:

There are two equations with three unknowns. Therefore, this system cannot have a unique solution, instead it will have an infinite number of solutions. Therefore, there cannot be a one-to-one mapping from

In addition to what Stephen Tashi said, your matrix A is incorrect. If T is a map from R³ to R², any matrix representation will have to be 2 X 3, not 3 X 2.

IOW, like this:
$$ A = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{bmatrix}$$

I would use $a_{ij}$ for the components of A, $x_1, x_2, x_3$ for the components of x, and $y_1, y_2$ for the components of y.

 

[_N3WTON_]

In addition to what Stephen Tashi said, your matrix A is incorrect. If T is a map from R³ to R², any matrix representation will have to be 2 X 3, not 3 X 2.

IOW, like this:
$$ A = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{bmatrix}$$

I would use $a_{ij}$ for the components of A, $x_1, x_2, x_3$ for the components of x, and $y_1, y_2$ for the components of y.

Thank you...so basically I should go back to the drawing board right? :D

 

[Mark44]

Back to the drawing board, but using the definition of "one-to-one transformation," which is something that Stephen alluded to.

 

[_N3WTON_]

I looked at this problem again this morning and I think my main problem was that I was writing the matrix as 3x2 instead of 2x3. With that in mind, I took a more simple approach (the approach I think the instructor intended us to take) and said:
A 2x3 matrix can have at most one pivot in each of its two rows. However, 3>2, so the same matrix cannot have a pivot in each of its three columns. Therefore, the given transformation can never be one-to-one.

 

[HallsofIvy]

I can't speak for your teacher's hint, but I would do it this way: Apply A to the standard basis for R³, u= (1, 0, 0), v= (0, 1, 0), and w= (0, 0, 1). The three vectors, Au, Av, and Aw can't be independent because they are in R² which has dimension 2. Therefore, there exist numbers x, y, and z such that xAu+ yAv+ zAw= 0. From that, zAw= -xAu- yAv so A(zw)= A(-xu- yv). Now show that zw is NOT equal to -xu- yv.

 

[_N3WTON_]

I can't speak for your teacher's hint, but I would do it this way: Apply A to the standard basis for R³, u= (1, 0, 0), v= (0, 1, 0), and w= (0, 0, 1). The three vectors, Au, Av, and Aw can't be independent because they are in R² which has dimension 2. Therefore, there exist numbers x, y, and z such that xAu+ yAv+ zAw= 0. From that, zAw= -xAu- yAv so A(zw)= A(-xu- yv). Now show that zw is NOT equal to -xu- yv.

Great, thank you for the hint!