T* is the adjoint of the operator T Let T be a linear operator on a finite dimen
ID: 1888264 • Letter: T
Question
T* is the adjoint of the operator T
Let T be a linear operator on a finite dimensional vector space V. Prove that (T*)* = T.Explanation / Answer
Below is the similar problem. try to solve from this one. ThanQ Let T be a linear operator on a finite-dimensional vector space V, and let b (beta) be an ordered basis for V. Prove that l (lamda) is an eigenvalue of T if and only if l (lamda) is an eigenvalue of [T]b (beta). ==> if lamda is an eigenvalue of T , then there exists nonzero vector x in V such that Tx = lambda x, Since B(better using capital letter for basis) is an order basis of V, let B ={v1,v2,..,vn} (assume dim V = n) x = E aivi for scalars a1,a2,..,an (E means summation over i) i.e. xB = (a1,a2,..,an)^T (T means transpose) be a column vector(in F^n) Let [T]B = [Tij](nxn matrix) Since Tx = lambda x, expressing in the o.b. B, we have [Tx]B = [lambda x]B, So, we get [Tx]B =[T]B xB = lambda xB Also, x is nonzero vector implies xB is non-zero in F^n. This shows lambda is an eigenvector of [T]B.Related Questions
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