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Exercise 7.D.14
Answers
Proof. First, we will prove that . From Exercise 5 in section 7A we see that .
Suppose . Then . Thus for some . We can write for some and . Thus . But 7.7 shows that . Therefore for some and so . Hence .
The inclusion in the other direction is easy. We have
Therefore .
Since is diagonalizable (because it is self-adjoint), it follows that the number of nonzero singular values of equals the dimension of . Note that . Therefore , completing the proof. □