If #M = ((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))# and #A# is an invertible rational #5xx5# matrix which commutes with #M#, then is #A# necessarily expressible as #A = aM^4+bM^3+cM^2+dM+e# for some scalar factors #a, b, c, d, e#?
This #M# is the companion matrix of the polynomial #x^5+4x+2# . It satisfies:
#M^5+4M+2 = 0# and since #x^5+4x+2# is irreducible (over #QQ# ), this is its minimum polynomial.
Hence the identity matrix #I_5# with #M# generates a field of matrices all expressible in the form #aM^4+bM^3+cM^2+dM+e# .
This
#M^5+4M+2 = 0# and since#x^5+4x+2# is irreducible (over#QQ# ), this is its minimum polynomial.
Hence the identity matrix
1 Answer
Yes
Explanation:
Note that the powers of
#M^0 = ((1, 0, 0, 0, 0),(0, 1, 0, 0, 0),(0, 0, 1, 0, 0),(0, 0, 0, 1, 0), (0, 0, 0, 0, 1))#
#M^1 = ((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))#
#M^2 = ((0, 0, 0, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0))#
#M^3 = ((0, 0, -2, 0, 0), (0, 0, -4, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0))#
#M^4 = ((0, -2, 0, 0, 0), (0, -4, -2, 0, 0), (0, 0, -4, -2, 0), (0, 0, 0, -4, -2), (1, 0, 0, 0, -4))#
Notice in particular that the left hand column of:
#aM^4+bM^3+cM^2+dM+eI#
is:
#((e),(d),(c),(b),(a))#
So given an invertible matrix
#((a_11),(a_21),(a_31),(a_41),(a_51))#
we find that the matrix:
#A-(a_51M^4+a_41M_3+a_31M_2+a_21M+a_11I)#
has left hand column:
#((0),(0),(0),(0),(0))#
...so is not invertible.
If
#A=a_51M^4+a_41M^3+a_31M^2+a_21M+a_11I#