322 CHAPTER 12. SELF ADJOINT OPERATORS

10. Show directly that if A is an n× n matrix and A = A∗ (A is Hermitian) then all theeigenvalues are real and eigenvectors can be assumed to be real and that eigenvectorsassociated with distinct eigenvalues are orthogonal, (their inner product is zero).

11. Let v1, · · · ,vn be an orthonormal basis for Fn. Let Q be a matrix whose ith columnis vi. Show

Q∗Q = QQ∗ = I.

12. Show that an n × n matrix Q is unitary if and only if it preserves distances. Thismeans |Qv| = |v| . This was done in the text but you should try to do it for yourself.

13. Suppose {v1, · · · ,vn} and {w1, · · · ,wn} are two orthonormal bases for Fn and sup-pose Q is an n × n matrix satisfying Qvi = wi. Then show Q is unitary. If |v| = 1,show there is a unitary transformation which maps v to e1.

14. Finish the proof of Theorem 12.7.5.

15. Let A be a Hermitian matrix so A = A∗ and suppose all eigenvalues of A are largerthan δ2. Show

(Av,v) ≥ δ2 |v|2

Where here, the inner product is (v,u) ≡∑n

j=1 vjuj .

16. The discrete Fourier transform maps Cn → Cn as follows.

F (x) = z where zk =1√n

n−1∑j=0

e−i 2πn jkxj .

Show that F−1 exists and is given by the formula

F−1 (z) = x where xj =1√n

n−1∑j=0

ei2πn jkzk

Here is one way to approach this problem. Note z = Ux where

U =1√n



e−i 2πn 0·0 e−i 2π

n 1·0 e−i 2πn 2·0 · · · e−i 2π

n (n−1)·0

e−i 2πn 0·1 e−i 2π

n 1·1 e−i 2πn 2·1 · · · e−i 2π

n (n−1)·1

e−i 2πn 0·2 e−i 2π

n 1·2 e−i 2πn 2·2 · · · e−i 2π

n (n−1)·2

......

......

e−i 2πn 0·(n−1) e−i 2π

n 1·(n−1) e−i 2πn 2·(n−1) · · · e−i 2π

n (n−1)·(n−1)

Now argue U is unitary and use this to establish the result. To show this verifyeach row has length 1 and the inner product of two different rows gives 0. NowUkj = e−i 2π

n jk and so (U∗)kj = ei2πn jk.

17. Let f be a periodic function having period 2π. The Fourier series of f is an expressionof the form

∞∑k=−∞

ckeikx ≡ lim

n→∞

n∑k=−n

ckeikx

and the idea is to find ck such that the above sequence converges in some way to f . If

f (x) =

∞∑k=−∞

ckeikx