Math Investigatiions
Vectorss, Matrices and
a Transform
mations
Naame:
Vectors
Vector addition is easy algebbraically, (v1 , v2 )  ( w1 , w2 )  (v1  w1 , v2  w2 ) , but what doees it mean allgebraically??
mean for vector subtractioon?
Whaat does this m
Dot Producct
g
by
The dot product of two vectoors, v  (v1 , v2 ) and w  ( w1 , w2 ) is given
v·w  v1w1  v2 w2
wever, this is uninterestin
ng. Let us seee what it does geometricaally.
How
Let v  (v1 , v2 )  (| v | cos( ),| v | sin( ))) and w  ( w1 , w2 )  (| w | cos(  ),| w | sin(  )) . Then θ = β - ∝ and
v·w  v1w1  v2 w2
c )
| v | cos( ) | w | sin(  ) | v | sin( ) cos(
| v || w | (cos(( ) cos(  )  sin( ) sin((  ))
| v || w | cos(    )
| v || w | cos( )
Therrefore, if




  0, tthen v·w | v || w | .

  , then v·w  0.
2
0  

2

2
, then v·w  0.
    , then v·w  0.
BASIS Sccottsdale Math Investigations
Vectors, Matrices and Transformations
Also, since cosine is an even function, and if we do some algebraic manipulation, we get that
Name:
Dot product is

linear: If k is a real number (kv)·w  k (v·w)

distributive: u·(v  w)  u·v  u·w

commutative: v·w  w·v
Exercises
Exercise 1: Find the angle between the points (1, 2) and ( (
3
1
 1,   3)
2
2
Exercise 2: Let l be the line through (5,1) and (2,-3), and m be the line through (-2,5) and (2,2). What is the angle of
their intersection?
Matrices
Before we start this lesson, remember that the standard basis vectors in math are (1,0,0), (0,1,0), and (0,0,1) in 3dimensions, and (1,0) and (0,1) in two dimensions.
Now, please, consider the matrix
 2 1
A

1 3
This is a function from  2  {( x, y ) | x  } to  2  {( x, y ) | x  } . We dot each row in the matrix with the
column (x, y) to achieve the image of the vector (x, y) under the function.
 2 1  x   2 x  y 
A( x, y )  
   

 1 3  y   x  3 y 
2
A(1, 0)  
1
We can see how the function works by applying to standard basis vectors. For example,
2
A(0,1)  
1
1 1   

3 0   
1 0   

3 1   
Summary: The column vectors of a matrix are the images of the ______________ _________________
___________________.
BASIS Scottsdale Math Investigations
Vectors, Matrices and Transformations
Here are the standard basis vectors before A is applied to them Name:
Draw the image of the standard basis vectors after A is applied to them (5 tick marks = 1 unit)
What is the area of the parallelogram on the right side? _______________________
So, the _____________ of the matrix is the area of the parallelogram generated by the image of the vectors (0,1) and
(1,0). This gives us an idea of what the matrix is doing to the square that is created by the standard basis vectors.
BASIS Scottsdale Math Investigations
Vectors, Matrices and Transformations
For each of the following matrices;
Name:
1. Draw the image of the standard basis vectors
2. Compute the determinant
3. Summarize the action of the matrix on the standard basis vectors or the original square.
Matrix
Image of Standard Basis Vectors
 1 1
B

 1 1
 4 3
C

1 0

1
D
0

1
5

4
5 
2 4
E

 1 2 
BASIS Scottsdale Summary
Math Investigations
Vectors, Matrices and Transformations
Name:
Row Vectors:
Since the image of the vector v = (x, y) under the function matrix, A, is the dot product of all the rows with the vector
v. Then it follows that each coordinate of the image can give us the angle between the vectors in the rows and the
vector v.
Example:
 2 3  1 5 
1 1 1  0 

   
 5 

 26 
1
Tells us that (1, 1) is perpendicular to (1, -1) and that the angle between (1,1) and (2,3) is cos 
which is approximately 11 degrees.
The most important lesson to take from what we’ve learned about row vectors is that the vectors that the matrix sends
to zero are perpendicular to all the row vectors of the matrix.
Column Vectors:
These are the images of the standard basis vectors under the matrix function. They give us a quick glance as to what
the matrix is doing to the entire space.
Determinant:
This measures the change in area of the parallelogram generated by the standard basis vectors after they’ve been
inputted into the matrix function. The most important point to take from this lesson is that if the determinant is zero,
then the matrix collapsed a dimension, and, hence, cannot be an injective function.
The converse is also true.
This called be generalized to three dimensions! The determinant of three vectors in three space is the volume of the
parallelopiped they generate.
Fixed point: A fixed point by a matrix, A, is a vector v such that A(v) = v.
BASIS Scottsdale Math Investigations
Vectors, Matrices and Transformations
Name:
Linear Transformations and Matrices
Translations:

Have no fixed point
A translation 5 units upward and 3 units to the left is the same as a translation by the vector (5, -3).
Hence, the translation can be written by adding to each point, (x, y), on the plane the vector (5, -3).
 x   5   x  5
 y    3   y  3
    

Example: Write the transformation matrix for the translation below.
Rotations: The only fixed point is the center of rotation.
Most rotations are about the origin, in the counter-clockwise direction.
(0,1) (1,0) A rotation by Ө in the counter clockwise direction will send the standard basis vectors
(1, 0)  (cos( ),sin( ))
(0,1)  ( sin( ), cos( ))
So the matrix of a rotation of theta degrees rotated counter clockwise around the origin is given by
 cos( ) sin( ) 
  sin( ) cos( ) 


The angle of the intersection of any line and its image under a rotation by angle
BASIS Scottsdale is still .
Math Investigations
Vectors, Matrices and Transformations
Name:
Reflections across a line y=mx: The fixed line is the line of reflection.
θ A reflection in the line making and angle   arctan( m ) where m is the slope of the line with the x axis can be
represented by the matrix below.
cos(2 ) sin(2 ) 
A

 sin(2 )  cos(2 ) 
This is because the standard basis vectors
(1, 0)  (cos(2 ),sin(2 ))
(0,1)  (sin(2 ),  cos(2 ))
Geometrically: The image of a point under a reflection is the same distance to the line of reflection as the original
point.
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