eigenvalues and vectors related proof

• Oct 18th 2009, 01:01 PM
excellents
eigenvalues and vectors related proof
Hello!

Could someone help me to proof one stuff?

Say, I have A which is a matrix of a linear operator. λ in Spec(A) ( i.e. λ is eigenvalue of A ) x is an eigenvector for λ, and f(X) is a polynomial.

How to proof that x is also an eigenvector for f(A)?

• Oct 18th 2009, 02:34 PM
HallsofIvy
Quote:

Originally Posted by excellents
Hello!

Could someone help me to proof one stuff?

Say, I have A which is a matrix of a linear operator. λ in Spec(A) ( i.e. λ is eigenvalue of A ) x is an eigenvector for λ, and f(X) is a polynomial.

How to proof that x is also an eigenvector for f(A)?

Write the polynomial as [tex]f(A)]= a_nA^n+ a_{n-1}A^{n-1}+ \cdot\cdot\cdot+ a_1A+ a_nI[/itex]. Then $\displaystyle f(A)x= aA^nx+ a_{n-1}A^{n-1}x+ \cdot\cdot\cdot+ a_1Ax+ a_nIx$.

Before you go further note that Ix= x, $\displaystyle Ax= \lambda x$, $\displaystyle A^2x= A(\lambda x)$$\displaystyle = \lambda Ax= \lambda(\lambda x)= \lambda^2 x$, etc.
• Oct 19th 2009, 04:31 AM
excellents
Where do $\displaystyle A^n x = \lambda^n x$ come from? Is that Cayley–Hamilton theorem?
• Oct 19th 2009, 07:42 AM
tonio
Quote:

Originally Posted by excellents
Where do $\displaystyle A^n x = \lambda^n x$ come from? Is that Cayley–Hamilton theorem?

No, it is induction: it's easy to show that $\displaystyle Ax=\lambda x \, \Longrightarrow A^2x=\lambda^2 x$

Now by induction show for any natural n.

Tonio