[VA] SRC #007 - 2020 April 1st Ramblings

[Bernd Grubert]

(04-01-2020 06:52 PM)Valentin Albillo Wrote:  3) Now, here's another 2^nd-degree equation in X:

      X² + 2 X + 5 I = 0

with the caveat that this time X is not just some scalar value but a square matrix and I is the corresponding Identity matrix. Matricial equations can have any number of roots, including an infinity of them or none at all, and it would be nice to find some roots for this equation, if they do exist.

The solutions of the quadratic equation are -1+2*i and -1-2*i. Trivial square matrices that satisfy the equation are diagonal matrices D that contain the solutions as elements in the diagonal, e.g.:

[-1 + 2 i, 0 ]
[ 0, -1 - 2 i]

Multiplying those diagonal matrices D from the left with an invertible matrix M of the same dimensions and with its inverse from the right
X = M D M^-1
is also a solution of the quadratic equation.
This can be seen by multiplying the quadratic equation with M^-1 from the left and with M from the right:
M^-1 X X M + 2 M^-1 X M + 5 M^-1 I M =
M^-1 (M D M^-1) (M D M^-1) M + 2 M^-1 (M D M^-1) M + 5 M^-1 I M =
(M^-1 M) D (M^-1 M) D (M^-1 M) + 2 (M^-1 M) D (M^-1 M) + 5 M^-1 I M =
D² + 2 D + 5 I

by making use of the fact that M^-1 M = M^-1 I M = I.
By definition the result of this expression is 0 since D is a root of the quadratic equation.
Since D is a solution of the equation so is X = M D M^-1.
Therefore there exist an infinite number of solutions for this equation.

Edit: Added parentheses. Restated the conclusion to make it more clear.