Line 100: Line 100:
 
=='''2. COORDINATE SYSTEMS'''==
 
=='''2. COORDINATE SYSTEMS'''==
  
''Requires basic knowledge of basis matrices.''
+
=='''COORDINATES'''==
 +
 
 +
''Requires basic knowledge of augmented matrix, basis and row reduced echelon form.''
  
 
Suppose
 
Suppose
  
<math>S = [v_{1}, v_{2}, v_{3}, ... , v_{n}]</math>
+
<math>S =  
is a basis of n-dimensional vector space V, then every vector v in V can be expressed as:
+
\{v_{1}, v_{2}, v_{3}, ... , v_{n}
 +
\}</math>
 +
is a basis of vector space V, then every vector v in V can be expressed as:
 +
 
 +
<math>v = a_{1} v_{1} + a_{2} v_{2}+ a_{3} v_{3}+ ... + a_{n} v_{n}</math>
 +
 
 +
where <math>a_{1}, a_{2}, a_{3}, ... , a_{n}</math> are real numbers.
 +
 
 +
Hence, we can refer
 +
 
 +
<math>[v]_{s} = \begin{bmatrix}
 +
a_{1} \\
 +
a_{2} \\
 +
a_{3} \\
 +
\vdots \\
 +
a_{n} \end{bmatrix}</math>
 +
 
 +
as ''coordinate vector of v with respect to basis S''. The contents of the matrix are called ''coordinates of v with respect to S''.
 +
 
 +
Example:
 +
 
 +
Let S = {<math>v_{1}, v_{2}, v_{3}</math>} be an ordered basis of <math>R^{3}</math>, where
 +
 
 +
<math>v_{1} = \begin{bmatrix}
 +
1 \\
 +
0 \\
 +
1 \end{bmatrix}</math>
 +
<math>v_{2} = \begin{bmatrix}
 +
-1 \\
 +
2 \\
 +
-2 \end{bmatrix}</math>
 +
<math> v_{3} = \begin{bmatrix}
 +
0 \\
 +
1 \\
 +
0 \end{bmatrix}</math>
 +
 
 +
Find <math>[v]_{s}</math> for
 +
 
 +
<math>v = \begin{bmatrix}
 +
-2 \\
 +
7 \\
 +
-5 \end{bmatrix}</math>
 +
 
 +
Step 1:
 +
 
 +
Create an augmented matrix from <math>v_{1} , v_{2}, v_{3}</math> and v
 +
 
 +
The augmented matrices for the basis and v above is:
 +
 
 +
<math>\begin{bmatrix}
 +
1 & -1 & 0|& -2 \\
 +
0 & 2 & 1|& 7 \\
 +
1 & -2 & 0|& -5 \end{bmatrix}</math>
 +
 
 +
Step 2:
 +
 
 +
Find the row reduced echelon form of the matrix. This will give you the solution of <math>a_{1}, a_{2}, a_{3}</math>
 +
 
 +
The row reduced echelon form of the matrix above is:
 +
 
 +
<math>\begin{bmatrix}
 +
1 & 0 & 0|&1\\
 +
0 & 1 & 0|&3\\
 +
0 & 0 & 1|&1\end{bmatrix}</math>
 +
 
 +
From this, we can obtain <math>a_{1} = 1, a_{2} = 3, a_{3} = 1</math>
 +
 
 +
Therefore,
 +
 
 +
<math>[v]_s = \begin{bmatrix}
 +
1 \\
 +
3 \\
 +
1 \end{bmatrix}</math>
 +
 
 +
 
 +
=='''ISOMORPHISM'''==
  
 +
Suppose v and w are vectors in vector space V with a basis S = <math>
 +
\{
 +
v_{1}, v_{2}, ... , v_{n}
 +
\}</math>
  
 
Marlina Triesjayanti
 
Marlina Triesjayanti

Revision as of 21:01, 6 December 2010

Matrix Multiplication and Coordinate Systems

From a large topic of Linear Algebra, I will focus specifically on Matrix Multiplication and Coordinate Systems. This page is created to enhance the understanding of these subtopics for MA 265 students.

1. Matrix Multiplication

There are some properties that make matrix multiplication unique and different from other real number multiplication.


Basic Information: Dot Product

Dot product or inner product of

$ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \cdot \begin{bmatrix} d \\ e \\ f \end{bmatrix} = ad + be + fc $


In order to be able to multiply two or more matrices, those matrices MUST satisfy this requirement:

the row size of the first matrix must be equal to the column size of the second matrix.

Let A be your first matrix and B as you second matrix.

Let C be the result matrix.

To multiply A and B, we must follow the 'row-column' rule, that is, compute the dot product of the numbers of R ROW in matrix A with the numbers of C COLUMN in matrix B. The result will be located in your $ R^{th} row $ and $ C^{th} $ column.

Example:

Find the matrix multiplication of

$ \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \times \begin{bmatrix} 7 & 8 & 9 \\ 10 & 11 & 12 \end{bmatrix} $

Step 1:

Compute the dot product of the $ r^{th} $ row with $ c^{th} $ column.

For example: first row by first column

We get:

$ 1 \times 7 + 2 \times 10 = 27 $

For second row by first column:

We get:

$ 3 \times 8 + 4 \times 10 = 61 $

Repeat this step to obtain 6 other results for the matrix.

Step 2:

Enter the results you obtain in step 1 according to their position.

For example, 27 is obtained from dot product of first row and first column. Hence, it is located in first row and first column. This applies to other numbers you got in step 1.

Therefore the result is:

$ \begin{bmatrix} 27 & 30 & 33 \\ 61 & 68 & 75 \\ 95 & 106 & 117\end{bmatrix} $

NOTE: Matrix multiplication is NOT commutative. The result of $ A \times B $ is not the same as $ B \times A $

Interesting Fact: If you multiply a matrix with M columns and N rows by a matrix with N columns and P rows, the resulting matrix will have M columns and P rows.

Other link related to this topic on Rhea : [1] MA 351 (Elementary Linear Algebra)


2. COORDINATE SYSTEMS

COORDINATES

Requires basic knowledge of augmented matrix, basis and row reduced echelon form.

Suppose

$ S = \{v_{1}, v_{2}, v_{3}, ... , v_{n} \} $ is a basis of vector space V, then every vector v in V can be expressed as:

$ v = a_{1} v_{1} + a_{2} v_{2}+ a_{3} v_{3}+ ... + a_{n} v_{n} $

where $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers.

Hence, we can refer

$ [v]_{s} = \begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \\ \vdots \\ a_{n} \end{bmatrix} $

as coordinate vector of v with respect to basis S. The contents of the matrix are called coordinates of v with respect to S.

Example:

Let S = {$ v_{1}, v_{2}, v_{3} $} be an ordered basis of $ R^{3} $, where

$ v_{1} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $ $ v_{2} = \begin{bmatrix} -1 \\ 2 \\ -2 \end{bmatrix} $ $ v_{3} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $

Find $ [v]_{s} $ for

$ v = \begin{bmatrix} -2 \\ 7 \\ -5 \end{bmatrix} $

Step 1:

Create an augmented matrix from $ v_{1} , v_{2}, v_{3} $ and v

The augmented matrices for the basis and v above is:

$ \begin{bmatrix} 1 & -1 & 0|& -2 \\ 0 & 2 & 1|& 7 \\ 1 & -2 & 0|& -5 \end{bmatrix} $

Step 2:

Find the row reduced echelon form of the matrix. This will give you the solution of $ a_{1}, a_{2}, a_{3} $

The row reduced echelon form of the matrix above is:

$ \begin{bmatrix} 1 & 0 & 0|&1\\ 0 & 1 & 0|&3\\ 0 & 0 & 1|&1\end{bmatrix} $

From this, we can obtain $ a_{1} = 1, a_{2} = 3, a_{3} = 1 $

Therefore,

$ [v]_s = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} $


ISOMORPHISM

Suppose v and w are vectors in vector space V with a basis S = $ \{ v_{1}, v_{2}, ... , v_{n} \} $

Marlina Triesjayanti


Back to MA265 Fall 2010 Prof Walther

Back to MA265

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman