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| G Patel |
 Posted: Sat Oct 14, 2006 5:24 pm |
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Guest
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Why is this not a proof of Cayley-Hamilton theorem (that a square
matrix satisfies it's characteristic polynomial)?
Proof:
Let A be any square matrix.
characteristic polynomial is defined as det( A - x I) = 0
To see that A satisfies it, plug it in for x:
det(A - A I) = det( A - A) = det ( zero sq. matrix ) = 0
Therefore A satisfies it's characteristic polynomial.
My book says the proof of this theorem is too advanced for intended
audience, and mentions plugging A into the determinant level equation
is not good enough (but doesn't say why). |
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| Axel Vogt |
| Posted: Sat Oct 14, 2006 6:11 pm |
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Guest
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G Patel wrote:
[quote:0d7ec781e1]
Why is this not a proof of Cayley-Hamilton theorem (that a square
matrix satisfies it's characteristic polynomial)?
Proof:
Let A be any square matrix.
characteristic polynomial is defined as det( A - x I) = 0
To see that A satisfies it, plug it in for x:
det(A - A I) = det( A - A) = det ( zero sq. matrix ) = 0
Therefore A satisfies it's characteristic polynomial.
My book says the proof of this theorem is too advanced for intended
audience, and mentions plugging A into the determinant level equation
is not good enough (but doesn't say why).
[/quote:0d7ec781e1]
The objection is: substituting A for x in x*I gives A*I = A, yes.
But x*I means: x on the diagonal, so x*I is a Matrix over k[x],
so is A - x I, it is not a matrix with entries in k, your field.
Now what is det? C-H states it is the 0 matrix, not the 0 in k.
So one talkes about: evaluating a polynomial (?) det over a
non-commutative ring. But the char polynomial lives in k[x],
not in the matrices over it.
So the objection is: "What should substitution be?" |
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| Guest |
| Posted: Sat Oct 21, 2006 7:02 am |
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G Patel wrote:
[quote:a4cdec020f]Why is this not a proof of Cayley-Hamilton theorem (that a square
matrix satisfies it's characteristic polynomial)?
Proof:
Let A be any square matrix.
characteristic polynomial is defined as det( A - x I) = 0
To see that A satisfies it, plug it in for x:
det(A - A I) = det( A - A) = det ( zero sq. matrix ) = 0
Therefore A satisfies it's characteristic polynomial.
[/quote:a4cdec020f]
It is true that det (A - AI) = 0. But does det (A - AI) expand as a
polynomial? In fact it does, but you need prove that too.
[quote:a4cdec020f]My book says the proof of this theorem is too advanced for intended
audience, and mentions plugging A into the determinant level equation
is not good enough (but doesn't say why).[/quote:a4cdec020f] |
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