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?
1 Answer
It is false.
Explanation:
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
#{ (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