This is the question Suppose T (V), where V is a finite-dimensional vector space

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This is the question

Suppose T (V), where V is a finite-dimensional vector space. Fix a basis B = (v1,...,vn) on V. Let A = M.(T) be the matrix of T with respect to the basis B. If v V, we use the notation M(v) for the coordinate vector of v with respect to B. Consider the linear transformation S: Fn -rightarrow Fn given by S(x) = Ax. Let lambda F. Prove: A is an eigenvalue of T if and only if A is an eigenvalue of S. v V is an eigenvector for T corresponding to A if and only if M(v) is an eigenvector of S corresponding to A.

Explanation / Answer

(a) & (b) in the same time, since (b) => (a)

Let v in V an eigenvector for T corresponding to L.

Note M(v) = Pv for the change of basis matrix P (from B to natural basis of R^n )

Note T_0 be the matrix of T is the natural basis, we have also T_0 = P^(-1)AP => AP = PT_0

First we have T(v)=Lv, thus by definition T_0v = Lv.

Per defintion S(M(v)) = AM(v) = APv = PT_0v=LPv=LM(v).

So T(v)=Lv <=> S(M(v))=LM(v)

v is an eigenvector for T corresponding to L <=> M(v) is an eigenvector of S corresponding to L

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