How do you find a #2xx2# matrix #A# with rational coefficients such that #A^2+A+((1,0),(0,1)) = ((0,0),(0,0))# ?
4 Answers
Explanation:
By Cayley-Hamilton, the
I get
Explanation:
I used
I get the system (equations numbered in the obvious way)
#a^2+bc+a=-1#
#ab+bd+b=0#
#ac+cd+c=0#
#d^2+bc+d=-1#
Subtracting equation 4 from equation 1 gets us
The only real solution is
From equation 1 (0r eq 4) we now get
At this point, I can find no other restrictions on
So, I tried
Unless I made some silly error,
More generally,
If
then
Now if we add
A couple of possible solutions:
#A = ((-1/2, 1/2),(-3/2, -1/2))" "# or#" "A = ((-1/2, -1/2),(3/2, -1/2))#
Explanation:
Just in case you have not encountered the notation, the set of all rational numbers is denoted by
First consider matrices of the form
If you add, subtract, multiply or divide (i.e. multiply by the multiplicative inverse), such matrices then you end up with a matrix of the same form.
Matrices of the form
In other words, such matrices form a field. This field is isomorphic to
#a -> ((a,0),(0,a))#
which preserves the arithmetical structure.
Now consider the matrix:
#((0, 1),(-3, 0))#
See what happens if we square this matrix:
#((0, 1),(-3, 0))((0, 1),(-3, 0)) = ((-3,0),(0,-3))#
So
If we add this to our field of matrices, and add all other matrices required to make it closed under arithmetic operations, then we have the set of matrices of the form:
#((a, b),(-3b, a))" "# where#a, b in QQ#
This corresponds to the Complex number
Now:
#x^2+x+1 = (x+1/2)^2-(sqrt(-3)/2)^2#
#color(white)(x^2+x+1) = (x+1/2-sqrt(-3)/2)(x+1/2+sqrt(-3)/2)#
Hence zeros:
#x = -1/2+-sqrt(-3)/2#
which correspond to the matrices:
#((-1/2, 1/2),(-3/2, -1/2))" "# and#" "((-1/2, -1/2),(3/2, -1/2))#
Footnote
In fact any matrix of the following form will work:
#((0, k),(-3/k, 0))#
and hence there are a family of solutions to the original problem.
One solution is the companion matrix
Explanation:
I think I was taught this about 35 years ago and had forgotten it...
Companion matrix
Given a monic polynomial:
#x^n + a_(n-1)x^(n-1) + a_(n-2)x^(n-2) + ... + a_1 x + a_0#
The companion matrix is an
#C = ((0, 0,..., 0, -a_0), (1, 0,..., 0, -a_1), (0, 1,...,0, -a_2), (vdots, vdots,ddots,vdots,vdots),(0, 0,...,1, -a_(n-1)))#
Then
#C^n + a_(n-1)C^(n-1) + a_(n-2)C^(n-2) + ... + a_1 C + a_0 = 0#
For a monic quadratic polynomial:
#x^2+bx+c#
The companion matrix is:
#C = ((0, -c), (1, -b))#
So for our example:
#x^2+x+1#
#C = ((0, -1), (1, -1))#
Footnote
I was interested in more complicated examples such as:
#x^5+4x+2 = 0#
The companion matrix for this example is:
#((0, 0, 0, 0, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0))#