# If #M = ((0,0,-1),(1,0,-1),(0,1,0))# and #A# is an invertible rational #3xx3# matrix which commutes with #M#, then is #A# necessarily expressible as #A = aM^2+bM+cI_3# for some scalar factors #a, b, c#?

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This #M# is the *companion matrix* of the polynomial #x^3+x+1# . It satisfies:

#M^3+M+I_3 = 0# and this is its minimum polynomial.

Hence the identity matrix #I_3# with #M# generates a field of matrices all expressible in the form #aM^2+bM+cI_3# .

This *companion matrix* of the polynomial

#M^3+M+I_3 = 0# and this is its minimum polynomial.

Hence the identity matrix

##### 1 Answer

Yes...

#### Explanation:

Given:

#M = ((0,0,-1),(1,0,-1),(0,1,0))#

Then:

#M^2 = ((0,0,-1),(1,0,-1),(0,1,0))((0,0,-1),(1,0,-1),(0,1,0)) = ((0, -1, 0),(0, -1, -1),(1, 0, -1))#

Suppose

We have:

#AM = ((c, u, v), (b, w, x), (a, y, z))((0,0,-1),(1,0,-1),(0,1,0)) = ((u, v, -c-u), (w, x, -b-w), (y, z, -a-y))#

#MA = ((0,0,-1),(1,0,-1),(0,1,0))((c, u, v), (b, w, x), (a, y, z)) = ((-a, -y, -z), (c-a, u-y, v-z), (b, w, x))#

Equating the elements of

#u = -a#

#w = c-a#

#y = b#

#v = -y = -b#

#x = -a-y = -a-b#

#z = w = c-a#

So:

#A = ((c, -a, -b),(b, c-a, -a-b),(a, b, c-a))#

#color(white)(A) = a((0, -1, 0),(0, -1, -1),(1, 0, -1))+b((0, 0, -1),(1, 0, -1),(0, 1, 0))+c((1, 0, 0),(0, 1, 0),(0, 0, 1))#

#color(white)(A) = aM^2+bM+cI_3#