The Spectral Theorem, algebra homework help

This is a part of Operators on Inner Product SpacesCHAPTER. From book “
Linear Algebra

Done Right” Page: 221 & 222.

I need more details and explanation for this Theoremdepending on the book only.

I have attached a copy of Theorem and its proof from the book.

222
CHAPTER 7 Operators on Inner Product Spaces
Proof First suppose (c) holds, so T has a diagonal matrix with respect to
some orthonormal basis of V. A diagonal matrix equals its transpose. Hence
T = T*, and thus T is self-adjoint. In other words, (a) holds.
We will prove that (a) implies (b) by induction on dim V. To get started,
note that if dim V = 1, then (a) implies (b). Now assume that dim V > 1 and
that (a) implies (b) for all real inner product spaces of smaller dimension.
Suppose (a) holds, so T E L(V) is self-adjoint. Let u be an eigenvector
of T with || || = 1 (7.27 guarantees that T has an eigenvector, which can
then be divided by its norm to produce an eigenvector with norm 1). Let
U = span(u). Then U is a 1-dimensional subspace of V that is invariant
under T. By 7.28(c), the operator T\u1 € L(U+) is self-adjoint.
By our induction hypothesis, there is an orthonormal basis of Ut consist-
ing of eigenvectors of Tlul. Adjoining u to this orthonormal basis of U1
gives an orthonormal basis of V consisting of eigenvectors of T, completing
the proof that (a) implies (b).
We have proved that (c) implies (a) and that (a) implies (b). Clearly (b)
implies (c), completing the proof.
1
7.29 Real Spectral Theorem
Suppose F = R and T E L(V). Then the following are equivalent:
(a)
T is self-adjoint.
(b)
V has an orthonormal basis consisting of eigenvectors of T.
(c)
T has a diagonal matrix with respect to some orthonormal basis
of V.