Existence of Eigenvector in Linear Operator in Complex Vector Space

Here is some quick note for the existence of eigenvector for linear operator T  in complex vector space of dimension n. Consider any non-zero vector X in the space, we have the n+1 vectors X, T[X],T^2[X],T^3[X],\cdots,T^n[X] to be linearly dependent. Thus, there exists a_0,a_1,\cdots,a_n such that a_0+a_1T[X]+a_2T^2[X]+\cdots+a_nT^n[X]=0. If a_n is zero, we will pick the largest k, k\le n, such that a_k \neq 0. In any case, we can write
a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0 with a_k\neq 0.
Now consider the polynomial expression
a_0+a_1x+a_2x^2+\cdots+a_kx^k. We can find complex numbers c_1,c_2,\cdots,c_k such that

a_0+a_1x+a_2x^2+\cdots+a_kx^k=a_k (x-c_1)(x-c_2)\cdots(x-c_k). Easy to verify that

a_k(T-c_1 I)(T-c_2 I)\cdots(T-c_k I)[X]
=a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0.

For the above to be true, there must exist an l, 1\le l \le k, such that

(T-c_{l} I)\cdots(T-c_kI)[X]=0 but (T-c_{l+1} I)\cdots(T-c_k I)[X]\neq 0.  Define Y_l=(T-c_{l+1} I)\cdots(T-c_k I)[X], note that (T-c_{l} I )[Y_l]=0 with Y_l\neq 0. That is T[Y_l]=c_l Y_l and thus c_l and Y_l are precisely an eigenvalue and eigenvector of T[\cdot].

 

Leave a Reply

Your email address will not be published. Required fields are marked *