How Does a Polynomial Transformation Affect Eigenvectors and Eigenvalues?

[ptolema]

Homework Statement

Let T be a linear operator (T: V-->V) on a vector space V over the field F, and let g(t) be a polynomial with coefficients from F. Prove that if x is an eigenvector of T with corresponding eigenvalue λ, then g(T)(x) = g(λ)x. That is, x is an eigenvector of g(T) with corresponding eigenvalue g(λ).

Homework Equations

T(x)=λx

The Attempt at a Solution

I tried substituting λx directly into g, but that gave me an answer that I couldn't easily factor x out of to get g(T)(x) = g(λ)x. I don't know if there is a theorem regarding this, but I've hit a wall. Any suggestions?

 

[HallsofIvy]

I don't see how doing what you say wouldn't work directly.

Write g(x) as $g(x)= a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0$. Then, for any vector x, $g(T)v= a_nT^nx+ a_{n-1}T^{n-1}x+ \cdot\cdot\cdot+ a_1Tx+ a_0x$

The crucial point is that, because x is an eigenvector of T with eigenvalue $\lambda$, $Tx= \lambda x$, $T^2x= T(Tx)= T(\lambda x)= \lambda T(x)= \lambda(\lambda x)= \lambda^2 x$, to $T^nx= \lambda^n x$. To prove that rigorously, use induction on n.

 

[halez12]

No, the previous answer was completely incorrect. First, (T(x))^2 is the quantity T(x) squared, which is not, in general, equal to T(T(x)), which is the quantity T(x) evaluated in T. Unless T is the squaring function, (T(x))^2 doesn't equal T^2(x).

Unfortunately, as you have it written, there is not a proof for such an equality, as it does not hold. consider g=x^2+1. Then g(T(x))=(T(x))^2+1= a^2x^2+1, which is obviously not equal to (a^2+1)x. (Where a is the eigenvalue to x). Even if we consider the notion given in the last post, the +1 term throws off the calculation, because g(a)x=xa^2+x, and g(T(x))=xa^2+1.

 

[stringy]

Nah, HallsofIvy has it right. Let A be a matrix representation of T with respect to some basis. Then

g(A) = a_n A^n + ... + a_1 A + a_0 I.

It's the same sort of operator/matrix/polynomial idea as in the Cayley-Hamilton theorem. (EDIT: I am NOT saying that theorem is relevant here, it's just the same sort of "put matrix into a polynomial" idea.) Then if x and c an eigenvector and eigenvalue of A, we are to prove

g(A)x = g(c) x.

 

[HallsofIvy]

No, the previous answer was completely incorrect. First, (T(x))^2 is the quantity T(x) squared

No, it's not. We are working in a general vector space and multiplication (and so squaring) is not even defined. The only thing "$T^2(x)$" (I never wrote "$(T(x))^2$") could mean is T(T(x)).

, which is not, in general, equal to T(T(x)), which is the quantity T(x) evaluated in T. Unless T is the squaring function, (T(x))^2 doesn't equal T^2(x).

Unfortunately, as you have it written, there is not a proof for such an equality, as it does not hold. consider g=x^2+1. Then g(T(x))=(T(x))^2+1= a^2x^2+1, which is obviously not equal to (a^2+1)x. (Where a is the eigenvalue to x). Even if we consider the notion given in the last post, the +1 term throws off the calculation, because g(a)x=xa^2+x, and g(T(x))=xa^2+1.