| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hi I'm having some difficulty solving this problem. Could anyone help me
get started. Or show me how they would solve it.
Thanks Dennis
Let A be an n x n matrix with characteristic polynomial
n n n-1
f(t) = (-1) t + a t + ... + a t + a
n-1 1 0
Prove That A is invertable if and only if a .NE. 0
0
also prove
if A is invertable then
-1 n n-1 n-2
A = (-1/a0) [(-1) A + a A + ... + a I ]
n-1 1 n
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1220.1 | A sketch of the proofs | COOKIE::PBERGH | Peter Bergh, DTN 523-3007 | Fri Apr 13 1990 16:30 | 35 |
<<< Note 1220.0 by ARCANA::ESTRELLA >>>
-< linear algebra problem >-
>> Let A be an n x n matrix with characteristic polynomial
n n n-1
>> f(t) = (-1) t + a t + ... + a t + a
>> n-1 1 0
>> Prove That A is invertible if and only if a .NE. 0
>> 0
This one is simple: the roots of the characteristic polynomial are the
eigenvalues of A. A matrix is invertible iff (if and only if) it has no
non-zero eigenvalues.
Alternatively, the characteristic polynomial is f(t) = det(A-tI). f(0)=0 iff
det(A)=0.
>> also prove
>> if A is invertible then
>> -1 n n-1 n-2
>> A = (-1/a0) [(-1) A + a A + ... + a I ]
>> n-1 1 n
It is known that any matrix satisfies its own characteristic equation (f(t)=0).
(The easiest way to prove this is to make a coordinate transformation that makes
the matrix diagonal; the calculations then become the same as with scalars.)
Multiply the equation you have by A (from the right) and you get the
characteristic eqaution after a few simple manipulations.
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| 1220.2 | thanks | ARCANA::ESTRELLA | Sat Apr 14 1990 20:23 | 3 | |
thanks
Dennis
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