# MATH 33AH Lecture Notes - Lecture 3: Linear Map, Glossary Of Video Game Terms, Formula AtlanticExam

by OC2716258

Department

MathematicsCourse Code

MATH 33AHProfessor

AllStudy Guide

FinalThis

**preview**shows pages 1-2. to view the full**6 pages of the document.**Math 33A Discussion Example

Austin Christian

October 23, 2016

Example 1. Consider tiling the plane by equilateral triangles, as below.

Let vand wbe the orange and green vectors in this ﬁgure, respectively, and let B′=

{v,w}be the basis for R2formed by these two vectors. Let ube the purple vector.

(a) Sketch the vector 1

2B′

in the ﬁgure.

(b) Give the components of uwith respect to the basis B′.

(c) Let T:R2→R2be the linear transformation that rotates the plane counterclockwise

through an angle of 2π/3. Find the matrix B′[T]B′.

(d) Is the terminal point of the vector −1008

2016 B′

in the center of one of the red hexagons

or on an edge?

(Solution)

(a) Because this vector is given in the basis B′={v,w}, the desired vector is equal to

1v+ 2w, which is the blue vector in the following ﬁgure:

1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

(b) We see that we can get to the terminal point of uby following wand then following

−vtwice. Alternatively, we could follow −vtwice and then follow w. In either case,

we see that

u=w−2v=−2

1B′

.

(c) When we apply this rotation, vand ware carried to their dashed counterparts in the

following ﬁgure:

We see that T(v) = w−vand T(w) = −v, so

T1

0B′=−1

1B′

and T0

1B′=−1

0B′

.

This means that the matrix of Twith respect to B′is given by

B′[T]B′=−1−1

1 0 .

(d) Notice that the vector −1

2B′

lies in the center of a hexagon:

It is not diﬃcult to see that integer multiples of this vector will also lie in the centers

of various hexagons. Because our vector is such a multiple, it lies in the center of the

1008-th hexagon above the hexagon containing the origin.

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