Dr Euler and characteristic equations

  • Thread starter Bob3141592
  • Start date
  • #1
234
2
In "Dr. Euler's Fabulous Formula" by Paul Nahin, early in chapter 1 is discussed characteristic polynomials of a square matrix and the Cayley-Hamilton theorem, that any square matrix A satisfies its own characteristic equation. On page 21 it states p(lambda) = lambda^2 + a1*lambda + a2 = 0 and then goes on to say "suppose we divide lambda^n by lambda^2 + a1*lambda + a2. The most general result is a polynomial of degree n - 2 and a remainder of at most degree one." Apparently this is an important result, but isn't this also an invalid divide by zero? Am I misunderstanding something here? I checked out characteristic equations in Schaum's Outline on Matrices but it doesn't touch on this point. Can somebody explain this? Thanks.
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
That is not divide by zero. You're dividing a (definitely non-zero) polynomial into another. Remember to divide f by g in this sense means to find polys q and r with

f=qg+r

and deg(r)<deg(f).
 

Related Threads on Dr Euler and characteristic equations

  • Last Post
Replies
5
Views
809
Replies
3
Views
1K
Replies
3
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
614
  • Last Post
Replies
1
Views
968
  • Last Post
3
Replies
60
Views
35K
Replies
2
Views
1K
Replies
1
Views
2K
Top