Given #M = ((1, 1, 1), (0, 5, 5), (0, 0, 7))#, is it true that there is a non-zero second degree polynomial of which #M# is a root?

Question

Sadie Ackerman · Accepted Answer

It is false.

We can check directly as follows: Given: #M = ((1, 1, 1), (0, 5, 5), (0, 0, 7))# We find: #M^2 = ((1, 1, 1), (0, 5, 5), (0, 0, 7))((1, 1, 1), (0, 5, 5), (0, 0, 7)) = ((1, 6, 13), (0, 25, 60), (0, 0, 49))# Then: #aM^2+bM+cI# #= a((1, 6, 13), (0, 25, 60), (0, 0, 49)) + b((1, 1, 1), (0, 5, 5), (0, 0, 7)) + c((1, 0, 0), (0, 1, 0), (0, 0, 1))# #= ((a+b+c, 6a+b, 13a+b), (0, 25a+5b+c, 60a+5b), (0, 0, 49a+7b+c))# So if #aM^2+bM+cI = 0# then looking at the top row we find: #{ (a+b+c = 0), (6a+b = 0), (13a+b = 0) :}# Subtracting the second equation from the third we find: #7a = 0" "# hence #a=0# Then from the second equation: #b = 0# Then from the first equation: #c = 0# That is #a=b=c=0#. So there is no non-zero 2nd degree polynomial of which #M# is a root.

Theodore Amidon · Answer

undefined Yes, it is true. Since the matrix M has a non-zero determinant (det(M) = 35), it implies that there exists a non-zero second-degree polynomial for which M is a root, based on the Cayley-Hamilton theorem.