How can I finish this proof and prove the converse?(linear algebra, orthogonality and diagonalization Qt...?

a) Since B is symmetric, B = P⁻¹DP for some orthogonal P and diagonal matrix D (with real entries; the eigenvalues, on the main diagonal). This is part of the "Spectral Theorem"

Then, A = B² = P⁻¹D²P.

Since A and D² share the same eigenvalues (being similar), and the diagonal entries (i.e., the eigenvalues) of D² are clearly non-negative, we conclude that the eigenvalues of A are all non-negative.

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b) By hypothesis, A = P⁻¹DP, P is orthogonal, and D's entries are the non-negative eigenvalues of A (by hypothesis on A and the Spectral Theorem).

Let B = P⁻¹ √D P, where √D is a diagonal matrix whose diagonal entries are the positive square roots of the corresponding entries from D; as in part a, these form the eigenvalues for B.

By construction, A = B² since B² = (P⁻¹ √D P)² = P⁻¹DP = A.

Finally, B is symmetric, because

B^t = (P⁻¹ √D P)^t = P^t √D^t (P⁻¹)^t = P^t √D (P⁻¹)^t = P⁻¹ √D (P⁻¹)⁻¹ = P⁻¹ √D P = B.

Note how P being orthogonal (P⁻¹ = P^t) was used is a crucial manner above.

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I hope this helps!

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