Question

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?

Answer 1

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 `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.