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I am having difficulty in understanding the above proof from "Linear Algebra Done Right" by Sheldon Axler.

  1. The way I have understood induction is "if it is true for n=1 and n & n+1, then it is true for all cases". But I could not explicitly see this approach in the above proof. Could someone throw some light on this?
  2. In the text above, I understood the part that if u is an eigenvector, then there exists another eigenvector that is from the orthogonal complement space. But I did not get what Axler means by "Adjoining u to orthonormal basis of $U^\bot$ gives an orthonormal basis of V consisting of eigenvectors of T". All I could get was u is an eigenvector and there is another eigenvector orthogonal to it. But how does it prove these form a complete basis of V? The previous theorem used here (7.27) says if T $\in$ L(V) is a self-adjoint operator, then T has an eigenvalue. But does it have enough eigenvectors to form the basis of V? What have I misunderstood? What does "adjoining u to ..." mean here?
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  • $\begingroup$ Perhaps you didn't mean exactly what you wrote but bear in mind that what you've written in (1) is not the way induction works. $\endgroup$ Aug 12, 2018 at 5:36

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  1. The first part of the induction ($n=1$) takes place in the second paragraph. The second part (if it holds for $n$, then it holds for $n+1$) in paragraphs 3 and 4.
  2. At this point of the proof, we're assuming that (b) holds for $U^\perp$. So, $U^\perp$ has an orthonormal basis consisting of eigenvalues of $T|_{U^\perp}$. Adding $u$ to this basis gives us an orthogonal basis of $V$ consisting of eigenvalues of $T$.
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  • $\begingroup$ dim (U's orthogonal complement) < dim V. Since we have assumed all spaces with smaller dimensions have orthonormal basis with eigenvectors, U's orthogonal complement also has them (since it is also self-adjoint). =>This is for dim V = n. A direct sum of of the basis of (U's orthogonal complement) with the eigenvector of U proves it for dim V = n+1. Have I reasoned it correctly? $\endgroup$ Aug 12, 2018 at 10:01
  • $\begingroup$ @SatheeshPaul Yes, you have. $\endgroup$ Aug 12, 2018 at 11:37

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