The change depends on direction of the motion: Generating Eigenvalues from special matrices

"The change depends on direction of the motion"

This means: "It is possible, we move but there will not be any change or even we maybe lose everything."
"The change depends on direction of the motion." is a philosophic quote in which mathematics (Differential Calculus) applied it to find out the definition of gradient vector field.
For example, in Economics, USD is still an independent variable where many curves such as x = f ($), y = f ($), z = ($) … are determining the direction of motion overall function of
w = f (x, y, z …).
In designing strategic plan, maybe whole a company's strategy – making hierarchy depends on only a person or a department where this independent variable will change the company to a star or overturned (collapsed). In physics, the internal energy of the gas (except an ideal gas) depends on the pressure, volume and temperature (K = f (p, v, T)).

Nowadays, we can see many matrices as transformations and operators which are affecting on vectors.

The purpose of this article is, to introduce some special matrices in which you can easily generate all eigenvalues without any calculation.

The question is: How can we apply these special matrices as operators in our real world?

Some special matrices

1. Let consider “A1” as set of Arithmetic Progression where:

d = 1,      a1 = 1   and    an = a1 + (n - 1) d, n = 1, 2, 3,…..

In this case, we have:

A1 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

One of permutations of set A1 is to invert members of set A1 as follows:

A2 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}      

We can generate many sets which are the periodicity of set A1 just like below cited:

A3 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

A4 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}

Finally, we will have set B:

B = {A1, A2, A3, A4, …….An}     


Rule 1: If size members of set A1 or A2 orA3 or An, is equal to size members of set B, we will have a square matrix (Mn*n) to be generated by sets A1, A2, A3,…….An in which Eigenvalue of this matrix will be calculated by using of Binomial Coefficient as follows:

Eigenvalue (Mn*n) = λ = C (k, 2)

Where:   k = n+1, nϵ N, λ > 0


Example:

A1 = {1, 2, 3, 4}

A2 = {4, 3, 2, 1}      

A3 = {1, 2, 3, 4}

A4 = {4, 3, 2, 1}  


Matrix (M 4*4) = 
   
1 2 3 4

4 3 2 1

1 2 3 4

4 3 2 1

λ = C ((4+1), 2) = 10

2. Let consider set “A” as follows:

A = {x | x ϵ R}

Rule 2: Each type square matrix which has been generated from set “A”  just like below matrix:

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

It will show us the eigenvalues which are just equal members set A which have been included in this square matrix (λ = a1, a2, a3, a4, …)


Example:

A = {0.67, 2, 43, 5, -23, 9, -2.3}

Assume we have set B which is a subset of A:

B = {43, -23, -2.3, 9)

Matrix N will be:

43   0  0  0  0

43  -23  0   0

43 -23 -2.3 0

43 -23 -2.3 9

Eigenvalues of Matrix N are: λ = 43, -23, -2.3, 9

3. Consider square matrices 2*2 as follows:

a    -a

b    -b

Or 

 a      b

-a    -b

a , b ϵ R

Rule 3: Eigenvalues of above matrices are equal to: λ = 0 and

λ = a – b

4. Consider square matrices 3*3 as follows:

 a     -a      a

 b     -b      b

 c     -c      c  

Or

  a     b       c

 -a    -b     -c

  a      b      c  

a , b , c ϵ R

Rule 4: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c)-b

5. Consider square matrices 4*4 as follows:

a      -a      a     -a

b     -b      b     -b

c     -c       c     -c

d     -d      d     -d

Or

  a     b      c      d

 -a    -b    -c     -d

 a      b      c      d

-a    -b     -c     -d

 a, b , c , d ϵ R

Rule 5: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c -d)-b

6. Consider square matrix “A” as follows:

            a       0      0

A =      a       b      0

            a       b      c  

If we turn this matrix around first column (as axis) and then we turn it around first row (as axis), we will have matrix “B”:

          c      b     a

B =    0      b     a

          0      0     a  

a , b , c ϵ R

Rule 6: Eigenvalue of A + B is equal to λ = c

and eigenvalue of A – B or B - A is equal to zero (λ = 0)

Example:

If matrix A is:

            1.67     0        0

A =      1.67     4        0

            1.67     4       -3  

            -3     4      1.67

B =       0      4     1.67

             0      0     1.67  

                    -1.33     4        1.67

A + B =        1.67      8       1.67

                     1.67      4      -1.33  

λ =  -3

                4.67     -4       -1.67

A - B =    1.67      0       -1.67

                1.67      4       -4.67  

λ = 0

7. Here is another special matrix. Consider matrix A is a square matrix n*n just like below forms in which (a1, a3, a5, a7 a9…an) are diagonal of matrix as follows:

A =

a1   a2    0     0     0
0    a3    a4    0     0
0    0      a5   a6    0
0    0      0     a7   a8
0    0      0      0   a9
…………………....an

Or

A =

a1     0    0    0    0

a2    a3    0    0    0

0     a4   a5    0    0

0     0    a6   a7    0

0     0    0    a8   a9

…………………....an

a1, a2, a3,….an ϵ R  

Rule 7: Eigenvalues of matrix A are all members on diagonal of matrix A. In fact, property of matrix A to generate eigenvalues is just like Diagonal Matrix.

Example:

A =

-1.56    2     0
  0     -12     3
  0        0     4  

λ = -1.56, -12 and 4

Consequently, we can find out that inverse of previous special matrix (Rule 7) will have the same eigenvalues (members of diagonal). According to Rule 7, we had below matrix M:

M =

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

Therefore, eigenvalues of M^-1 are all members on diagonal of M^-1.

8. Consider "T" as set of vectors as follows:

T = {V1, V2, V3, ….Vn}

Where:

V1 = a1i

V2 = a2 i + a3 j

V3 = a4 i + a5 j + a6 k

Vn = a7 i + a8 j + a9 k + a10 t …….. amtn   and     a1, a2, a3, a4, a5, …. am are members of R (real 

numbers)

Theorem 8:

If "A" is a square matrix n*n inferred from the set of above vectors just like below cited:

A = {(a1, 0, 0, 0,…), (a2, a3, 0, 0, 0, …), (a4, a5, a6, 0, 0, 0, …..), …....Vn}

In fact, Scalar amounts of V1, V2, V3, …Vn are respectively rows of matrix A.

Then:   Eigenvalues of (A^n) = all members on diameter of matrix A^n (n member of N).

A =

4	0	0	0	0
-6.5	1	0	0	0
5	2	7	0	0
3	9	-5	12	0
-3.45	9	43	15	72

A^4 =

256	0	0	0	0
-552.5	1	0	0	0
2210	800	2401	0	0
-13147.5	13995	-18335	20736	0
-249381	4740402	17264326	6713280	26873856

Eigenvalues are = 256, 1, 2401, 20736, 26873856