Cayley-Hamilton theorem for Operator

In summary, the Cayley-Hamilton theorem for operators states that every linear operator satisfies its own characteristic polynomial, which is useful in simplifying calculations and proving properties without explicitly finding eigenvalues and eigenvectors. This theorem is applicable to all types of linear operators and is a generalization of the Cayley-Hamilton theorem for matrices, but it cannot be extended to non-linear operators.
  • #1
zetafunction
391
0
let be [tex] f(x)=det(xI-T) [/tex] for some operator 'T'

then does Cayley-Hamilton theorem apply so [tex] f(T)=0 [/tex] in the sense of operator
 
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  • #2
Good question , Zetafunction.
I think we need restrictions on T to ensure convergence in the corresponding Banach/Hilbert space( or else f(T) may not make sense.)
 

1. What is the Cayley-Hamilton theorem for operators?

The Cayley-Hamilton theorem for operators states that every linear operator satisfies its own characteristic polynomial, meaning that if we plug the operator into its own characteristic polynomial, the resulting polynomial will be equal to the zero operator.

2. How is the Cayley-Hamilton theorem for operators useful?

The theorem is useful in many areas of mathematics, including linear algebra and differential equations. It allows us to simplify calculations and prove certain properties of operators without having to explicitly find their eigenvalues and eigenvectors.

3. Is the Cayley-Hamilton theorem applicable to all types of operators?

Yes, the theorem is applicable to all types of linear operators, including square matrices, differential operators, and integral operators.

4. How is the Cayley-Hamilton theorem related to the Cayley-Hamilton theorem for matrices?

The Cayley-Hamilton theorem for operators is a generalization of the Cayley-Hamilton theorem for matrices. The matrix version is a special case where the operator is a square matrix.

5. Can the Cayley-Hamilton theorem be extended to non-linear operators?

No, the Cayley-Hamilton theorem only applies to linear operators. Non-linear operators do not have characteristic polynomials, and therefore cannot satisfy the theorem.

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