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Homework Help Overview

The discussion revolves around transformations in linear algebra, specifically focusing on rotations and reflections. Participants are exploring how to represent these transformations using matrices and the implications of combining them.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to find a matrix representation for a transformation that involves a rotation followed by a reflection. There are questions about how to combine these transformations and what the resulting matrices would look like. Some participants are also discussing the mapping of standard unit vectors in relation to linear transformations.

Discussion Status

The discussion is active, with participants sharing their reasoning and calculations. Some have provided insights into how to derive the resulting matrices from the transformations, while others are questioning the assumptions made about the transformations and their effects on the vectors.

Contextual Notes

There are mentions of specific transformations (e.g., rotation through pi, reflection in the X axis) and how they interact, but some participants express confusion about the outcomes and seek clarification on systematic approaches to transformations.

stunner5000pt
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In each case find a rotation or reflection which equals the given transformation

Rotation through pi followed by reflection in the X axis

Now is there a way to actually work this out?

The text says the answer is simply reflection on the Y axis.
I tried to visualize by drawing a vector, rotating it pi and then reflecting it on the X axis but it doesn't seem to make sense...

A rotation thru pi would yield a matrix like this
[tex]\left(\begin{array}{cc} -1&0 \\ 0&-1 \end{array}\right)[/tex]How would one incorporate the reflection on the X Axis.

For that matter, how would one incorporate a Y axis reflection or a line y = mx reflection onto a matrix?
 
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if you reflect over the x axis, it only effects the y's. specifically it makes them negative. so that matrix would be
[tex]\left(\begin{array}{cc} 1&0 \\ 0&-1 \end{array}\right)[/tex]

when you multiply the two matrices, you get:
[tex]\left(\begin{array}{cc} -1&0 \\ 0&-1 \end{array}\right) \left(\begin{array}{cc} 1&0 \\ 0&-1 \end{array}\right) = \left(\begin{array}{cc} -1&0 \\ 0&1 \end{array}\right)[/tex]

when you reflect over the y axis, then you're just making all the x's negative. which, is the matrix we got as our answer.
 
Here is a neat trick. In any linear transformation, you only need to know where the standard unit vectors go. Let's say you have a linear transformation, and you know from the definition of this transformation that the vector
[tex]\left(\begin{array}{cc} 1 \\ 0 \end{array}\right)[/tex]
should go to
[tex]\left(\begin{array}{cc} 0.5 \\ 0.2 \end{array}\right)[/tex]

Now you have a matrix
[tex]\left(\begin{array}{cc} a&b \\ c&d \end{array}\right)[/tex]
Now what could a and c possibly be so that (1, 0) gets mapped to (0.5, 0.2)? Think about how to multiply a vector by a matrix.
 
Last edited:
Orthodontist gave you exactly what you need to know. (1, 0) is rotated 180 degrees counter clockwise so it ends up where? Reflecting that in the x-axis changes it to? That's your first column.
(0,1) is rotated 180 degrees clockwise so it ends up where? Reflecting that in the x-axis changes it to? That's your second column.
 
0rthodontist said:
Here is a neat trick. In any linear transformation, you only need to know where the standard unit vectors go. Let's say you have a linear transformation, and you know from the definition of this transformation that the vector
[tex]\left(\begin{array}{cc} 1 \\ 0 \end{array}\right)[/tex]
should go to
[tex]\left(\begin{array}{cc} 0.5 \\ 0.2 \end{array}\right)[/tex]

Now you have a matrix
[tex]\left(\begin{array}{cc} a&b \\ c&d \end{array}\right)[/tex]
Now what could a and c possibly be so that (1, 0) gets mapped to (0.5, 0.2)? Think about how to multiply a vector by a matrix.
so then i would have to multiply

[tex]\left(\begin{array}{cc} 1 \\ 0 \end{array}\right)^T \left(\begin{array}{cc} a&b \\ c&d \end{array}\right) = \left(\begin{array}{cc} 0.5 \\ 0.2 \end{array}\right)[/tex]

right?

so what if (1,0) was rotated pi/2? I know it would be (0,1)... but ist here a systematic way of doing it? What about pi/4?
 

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