Let V be a finite-dimensional vector space over the field F.

Let T be a linear operator on V.

Let p be the minimal polynomial for T.

Let c be a scalar in F.

Prove that
p(c) =0 if and only if c is an eigenvalue of T.


Suppose p(c) =0. p= (x-c)q for some q in F[x].
Since p is the unique monic generator of a nonzero ideal M = { p in F[x] : p(T) = 0} in F[x], q is not in M. This implies q(T) != 0 and there exists w in V s.t. q(T)(w) != 0. Let v = q(T)(w). Since p is in M, 0= p(T)(w). 0= p(T)(w)= (T-cI)(q(T)(w)) = (T-cI)(v).
Thus, c is an eigenvalue of T and v is an eigenvector of T associated with c.

Suppose c is an eigenvalue of T. Let v be an eigenvector of T associated with c. Tv = cv, v!=0.
Since p is in M, p(T) =0, p(T)v = p(c)v = 0. Since v !=0, p(c) = 0.






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