The Spectral Theorem, algebra homework help

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This is a part of Operators on Inner Product Spaces CHAPTER. From book "Linear Algebra

Done Right" Page: 221 & 222.

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

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

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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 โ‚ฌ 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.
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Explanation & Answer

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Suppose F = R and T E L(V)

A: T is self-adjoint
B: V has orthonormal basis consisting of eigenvectors of T
C: T has a diagonal matrix with resp...


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