Wednesday, October 30, 2013

Matrix Calculations: det(A - B * λ)=0

Let A and B be square matrices. B is a diagonal matrix. A diagonal matrix is a matrix with entries in it's diagonal positions, (1,1), (2,2), and so on, and 0 everywhere else. The identity matrix is a diagonal matrix. Another example is:

[ [2, 0, 0],[0, -2, 0],[0, 0, 5] ]

Let m, n, r, s, and t be numerical constants. Let a1, a2, and a3 be entries on the first row. Similar for b1, b2, b3, c1, c2, and c3.



det(A - B * ident(λ)) = 0



Comparison of 2 x 2 Matrices:
Note the constant term (λ^0), typed in green, the same.

B is a 2 x 2 identity matrix.

det( [ [a1, a2],[b1, b2] ] - λ * [ [1,0],[0,1] ] ) = 0
det( [ [a1, a2],[b1, b2] ] - [ [λ, 0],[0, λ] ]) = 0
λ^2 - (a1 + b2) * λ + a1*b2 - a2*b1 = 0

B is a diagonal matrix [ [m, 0],[0, n] ].

det( [ [a1, a2],[b1, b2] ] - λ * [ [m,0],[0,n] ] ) = 0
det( [ [a1, a2],[b1, b2] ] - [ [m*λ, 0],[0, n*λ] ]) = 0
m*n*λ^2 - (a1*n + b2*m)*λ + (a1*b2 - a2*b1) = 0



Comparison of 3 x 3 Matrices:
Note the constant term (λ^0), typed in green, the same.

B is a 3 x 3 identity matrix.

det( [ [a1,a2,a3],[b1,b2,b3],[c1,c2,c3] ] - λ * [ [1, 0, 0],[0, 1, 0],[0, 0, 1] ] ) = 0
det( [ [a1,a2,a3],[b1,b2,b3],[c1,c2,c3] ] - [ [λ, 0, 0],[0, λ, 0],[0, 0, λ] ] ) = 0

- λ^3
+ λ^2 * (a1 + b2 + c3)
+ λ * (-a1*b2 - a1*c3 - b2*c3 + a2*b1 + a3*c1 + b3*c2)
+ (a1*b2*c3 - a1*b3*c2 - a3*b2*c1 - a2*b1*c3 + a2*b3*c1 + a3*b1*c2) = 0

B is a general 3 x 3 diagonal matrix, [ [r, 0, 0],[0, s, 0],[0, 0, t] ].

det( [ [a1,a2,a3],[b1,b2,b3],[c1,c2,c3] ] - λ * [ [r, 0, 0],[0, s, 0],[0, 0, t] ] ) = 0
det( [ [a1,a2,a3],[b1,b2,b3],[c1,c2,c3] ] - [ [r*λ, 0, 0],[0, s*λ, 0],[0, 0, t*λ] ] ) = 0

- λ^3 * r*s*t
+ λ^2 * (s*t*a1 + r*t*b2 + r*s*c 3)
+ λ * (a1*b2*c3 - a1*b3*c2 - a3*b2*c1 - a2*b1*c3 + a2*b3*c1 + a3*b1*c2) = 0


Of course to get the eigenvalues (λ), solve the appropriate polynomial.

Hope this is helpful to those studying linear algebra or for those who are curious. This post was inspired from a program I was helping a fellow HP Prime programmer.

This post is dedicated to Michael de Estrada.


This blog is property of Edward Shore. 2013

2 comments:

  1. Great post. I was pleasantly surprised to learn that the determinate of a 3x3(with x coords in column 1, y coords in column 2, and 1's in column 3) is twice the area of the triangle that the coordinates form.

    ReplyDelete
    Replies
    1. Cool, Jason. I didn't even recognize that.

      Delete

Next Week... and Plans for October 2017

I'm so excited, can't want for next week's HHC 2017 calculator conference in Nashville!  It is my annual calculator conference ...