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cool matrix (5)

Finally I would like to talk about cool matrices.

Let's think about a three dimensional vector [x1 x2 x3]' =


x1
[x2]
x3.


(Here ' means transpose.) The vector of difference of each term is [(x1 - 0) (x2 - x1) (x3 - x1)]'. Let's think about a matrix A that makes this difference in the right hand side. I will tell you why we think such a matrix. A is


[ 1 0 0
-1 1 0
0 -1 1].


Let's check it out.


[ 1 0 0 [x1 [ x1 - 0
-1 1 0 x2 = x2 - x1
0 -1 1] x3] x3 - x2 ]


For instance, use [x1 x2 x3]' = [1 4 9]',


[ 1 0 0 [1 [ 1
-1 1 0 4 = 3
0 -1 1] 9] 5 ].


The right hand side is the difference of each term. By the way, this matrix has its inverse matrix B,


[1 0 0
1 1 0
1 1 1].


Let's compute A B,


[ 1 0 0 [1 0 0 [ 1 0 0 [1 0 0
-1 1 0 1 1 0 = (-1+1) 1 0 = 0 1 0
0 -1 1] 1 1 1] (-1+1) (-1+1) 1] 0 0 1].


Therefore, A^{-1} = B. Please be patience a bit more. Let's look into what is B.


[ 1 0 0 [x1 [x1
Bx = 1 1 0 x2 = x1 + x2
1 1 1] x3] x1 + x2 + x3]


You see B is adding terms. A does difference and B does sum. I was surprised here. The idea ``difference'' belongs more to calculus, compare to linear algebra. In calculus, difference is d/dx. The inverse operation of difference is integral (\int). These matrices are:

A ... d/dx
B ... \int

and we know A^{-1} = B.

A^{-1} = {d/dx}^{-1} = B = \int
===> {d/dx}^{-1} = \int

Matrix behaves an operator, now we think matrix A is a differential operator and B is an integral operator, then these are inverse operation each other. I hope you can see why I surprised. We can see an analogy of the fundamental theorem of calculus in a matrix operation. So I think these matrix A and B are cool matrices. I might be only one to think so, I can imagine many people don't agree with me... Well.


This matrix story is in the book: Gilbert Strang, ``Introduction to LINEAR ALGEBRA''. This is my personal hot book. He describes what is the inverse of matrix with calculus analogy. That is fantastic and I like his book.

For the people who is lazy as me, here is a octave code.


--- BEGIN diff_and_int.m ---
a = [1 0 0; -1 1 0 ; 0 -1 1]
inv(a)
--- END diff_and_int.m ---


When you run this code with typing 'octave diff_and_int.m', you will see
the matrix B.

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