Let
be an inner product space (over
or
), and
. Prove that
is an eigenvalue of
if and only if
(the conjugate of
) is an eigenvalue of
(the adjoint of
).

Let be an inner product space (over or ), and . Prove that is an eigenvalue...
Let
, and let
be a polynomial. Show that if is an
eigenvalue of , then is an
eigenvalue of .
Hint: this follows from the more precise statement that if
is a
non-zero eigenvector for for the eigenvalue
, then is also an
eigenvector for for the
eigenvalue . Prove
this.
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Let V be a finite dimensional inner product space,
w1,w2V. Let
TL(V)
and Tv=<v,w1>w2 for all vV.
Find all eigenvalues and the corresponding eigenspaces of T. Please
provide full solution.
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suppose
prove that 0 is the only eigenvalue of N
(hint: fist show 0 is an eigenvalue of N, and then show if
is any
eigenvalue then =0
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(5) Prove or give a countcrcxample: If A, B E Cnx"are sclf-adjoint, then AB is also self-adjoint. (6) Let V be a finitc-dimensional inner product space over C, and suppose that T E C(V) has the property that T*--T (such a map is called a skew Hermitian operator (a) Show that the operator iT E (V) is self-adjoint (i.c. Hermitian) (b) Prove that T has purely imaginary eigenvalues (i.e. λ ίμ for μ E R). (c) Prove that T has...
Let (V, (,) be an inner product space and T:V → V and S:V + V be self adjoint linear transformations. Show that TOS:V + V is self adjoint if and only if SoT=TOS.
Let V be a Hilbert space. Let S1 and S2 be two hyperplanes in V defined by Let be given. We consider the projection of y onto , i.e., the solution of (1) (a) Prove that is a plane, i.e., if , then for any . (b) Prove that z is a solution of (1) if and only if and (2) (c) Find an explicit solution of (1). ( d) Prove the solution found in part (c) is unique. We...
Q6 5 Points Let (V, (,)) be an inner product space and T :V + V and S: V + V be self adjoint linear transformations. Show that To S:V + V is self adjoint if and only if S T =To S.
Let V be a finite-dimensional inner product space. For an operator TEL(V), define its norm by ||T|:= max{||Tull VEV. ||0|| = 1}. (1) To explain this, note that {l|Tu ve V, || 0 || = 1} is a non-empty subset of [0,00). The expression max{||TV|| | V EV, ||0|| = 1} means the maximum, or largest, value in this set. In words, the norm of an operator describes the maximal amount that it 'stretches' (or shrinks) vectors. (a) (1 point)...
Let
be a metric space and let
be the topology on
induced by
, and let
be a compact space. Prove that
is compact.
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Prove that a normal operator on a complex inner product space is self- adjoint if and only if all its eigenvalues are real. [The exercise above strengthens the analogy (for normal operators) between self-adjoint operators and real numbers.]