Linear Transformations: Understanding n and m in T: R^n -> R^m

[KongMD]

Homework Statement

Given A =
<br /> \left(\begin{array}{ccc}1&amp;-1&amp;1\\0&amp;1&amp;1\end{array}\right)<br />

Why isn't Latex working for above array :(

Define a transformation as

T: \Re^{n} -&gt; \Re^{m}
T(\vec{x}) = A \vec{x}

1)
a. What is n?
b. What is m?

2) Find \vec{x} , if possible, given that

T(\vec{x}) = \left(\begin{array}{c}-1\\0\end{array}\right)

Homework Equations

The Attempt at a Solution

1) From my notes, the superscript of the first R should indicate the columns in matrix A and the superscript of the second R should indicate the rows in matrix A. If I am doing this correctly:

n = 3
m = 2

However, I really have lost track of what R means, here. I know it's "the set of all real numbers", but I don't really understand how you can convert something from "threespace" to "twospace".

2)

<br /> \left(\begin{array}{c}-1\\0\end{array}\right) = \left(\begin{array}{ccc}1&amp;-1&amp;1\\0&amp;1&amp;1\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\end{array}\right)<br />

 

[Robert1986]

A matrix is just a function that goes from R^n to R^m. But what is a function? A function from a set D(f) to a set R(f) is a rule that assigns a value in R(f) to each element in D(f). This is all the matrix is doing. It is taking a value in R^3 and assigning it a value in R^2. The R just means that the entries of the vectors in R^n and R^m are real numbers.

 

[KongMD]

So, was I correct in stating n = 3 and m = 2?

Also, I'm still stumped on problem #2. I don't think the Latex formatting came out right in my "attempt at a solution", so I'll post the work I've done on it here. I don't think the syntax is working, so I might just have to scan the physical page and upload it as an image.

EDIT: I guess I'll try and type this, though it'll probably take way longer. Double-slashes indicate new row.

2) T([-1\\0]) = [1 -1 1 \\ 0 1 1] [x₁ \\ x₂ \\ x₃]

 

[squaremeplz]

So you are mapping from the 3 dimensional real space to the 2 dimensional real space?

And you want to find the vector x?

2 is wrong.

A * x = [-1;0]

solve for x

 

[Mark44]

Fixed your LaTeX. Your [ tex] and [ /tex] tags were in the wrong places.

Homework Statement

Given A =
[[1 -1 1]
[0 1 1]]

Define a transformation as

T: \Re^{n} -&gt; \Re^{m}
T(\vec{x}) = A \vec{x}

1)
a. What is n?
b. What is m?

2) Find \vec{x} , if possible, given that T(\vec{x}) =
[[-1
[ 0]]

Does this format okay for you guys?
\left(\begin{array}{c}-1\\0\end{array}\right)

Homework Equations

The Attempt at a Solution

1) From my notes, the superscript of the first R should indicate the columns in matrix A and the superscript of the second R should indicate the rows in matrix A. If I am doing this correctly:

n = 3
m = 2

However, I really have lost track of what R means, here. I know it's "the set of all real numbers", but I don't really understand how you can convert something from "threespace" to "twospace".

2)

\left(\begin{array}{c}-1\\0\end{array}\right) = \left(\begin{array}{ccc}1&amp;-1&amp;1\\0&amp;1&amp;1\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\end{array}\right)

 

[KongMD]

Thanks for fixing up my Latex formatting. I tried fixing more of my original post so it would be cleaner with the Latex, but I just ended up mucking it up more. I've been doing Linear Algebra and Calc II homework for the past 4 hours and my brain is fried. I have the start and end tags in the right place, so what's the problem?

Oh well. I'll ask the professor tomorrow morning about number 2 and edit some of these posts so that this topic isn't a lost cause, if I figure it out. Thanks for helping out, though :)

 

[Mark44]

Write the matrix equation Ax = b as an augmented matrix, then row-reduce the matrix.
\left[\begin{array}{ccccc}1&amp;-1&amp;1&amp;|&amp;1\\0&amp;1&amp;1&amp;|&amp;0\end{array}\right]

There is not a unique solution for x.

 

[KongMD]

I thought as much. I actually had that written down as the work for my problem, but was confused because the RREF'd solution matrix doesn't include a value for x₃, since it only has two rows. Does this mean that x₃ is "free" and therefore there's no unique solution?

 

[Mark44]

Yes. To get a better understanding of this problem, it helps to look at the geometry of what the augmented matrix I wrote in post #7 represents. This matrix represents the system
x - y + z = 1
0x + y + z = 1

Each equation represents a plane in space. The system could represent one of three possibilities:
1) The planes are parallel and don't intersect at all.
2) The planes are parallel and coincide at each point.
3) The planes intersect in a line.
By row-reducing the augmented matrix you found that there was a free variable, which means that the planes intersect in a line.

Other systems of equations could produce the other two situations. For example, this system represents parallel planes with no points in common.
x - y + z = 1
x - y + z = 2

Row reduction would produce this augmented matrix.
\left[\begin{array}{ccccc}1&amp;-1&amp;1&amp;|&amp;1\\0&amp;0&amp;0&amp;|&amp;1\end{array}\right]
The second row represents the equation 0x + 0y + 0z = 1, which is not possible.


This system represents two equations for the same plane.
x - y + z = 1
2x - 2y + 2z = 2
Row reduction would produce this augmented matrix.
\left[\begin{array}{ccccc}1&amp;-1&amp;1&amp;|&amp;1\\0&amp;0&amp;0&amp;|&amp;0\end{array}\right]
The second row places no restrictions on the variables, so every solution of the first equation is also a solution of the second.

 

[KongMD]

Thanks a ton, Mark! It's so helpful to be able to just sit down at the end of a busy day and absorb this material at my own pace. I try and get help from my professor regarding these types of problems, but sometimes those meetings are rushed, and the concepts don't stick as much. Other times our schedules clash, and I have to depend on my professor to check his email and respond (which sometimes doesn't happen before the next class).

I really appreciate the help I've gotten on these forums. The replies I get never do the problem for me, but push me towards the right solution. Perhaps one day, when I'm not taking 15 credits and working a job, I will come back and "Pay It Forward" to the next generation of students who have questions.