# 2 proofs for unitary matrix/transformation prob.

• February 17th 2011, 06:36 PM
hashshashin715
2 proofs for unitary matrix/transformation prob.
first one is if A and B are unitary, then prove (or disprove) A + B is unitary.

so we have AA* = I and BB* = I.

AA* + BB* = 2I

Can you do this: (A + B)* = (A* + B*)?
If you can, then we can have (A + B)(A + B)* = AA* + AB* + BA* + BB* = I
so we would only need to prove AB* + BA* = -I. How do you do that?

2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.
• February 17th 2011, 06:51 PM
tonio
Quote:

Originally Posted by hashshashin715
first one is if A and B are unitary, then prove (or disprove) A + B is unitary.

Both $I\,,\,-I$ are unitary, so...

so we have AA* = I and BB* = I.

AA* + BB* = 2I

Can you do this: (A + B)* = (A* + B*)?
If you can, then we can have (A + B)(A + B)* = AA* + AB* + BA* + BB* = I
so we would only need to prove AB* + BA* = -I. How do you do that?

2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.

Prove that, in general, if $\lambda$ is an eigenvalue of $A$ , then $\overline{\lambda}$ is an

eigenvalue of $A^*$ (we assume all through that we use an orthonormal basis)

Tonio
• February 18th 2011, 01:50 AM
Ackbeet
For your first proof, note that

$A=\begin{bmatrix}1 &0\\ 0&1\end{bmatrix}$ and

$B=\begin{bmatrix}0 &1\\ 1&0\end{bmatrix}$ are both unitary (check this!).

But

$A+B=\begin{bmatrix}1 &1\\ 1&1\end{bmatrix}$ is not unitary (why not?).
• February 18th 2011, 02:20 AM
FernandoRevilla
Quote:

Originally Posted by hashshashin715
2nd proof:We have a real orthogonal n x n matrix A. If lamda is a complex eigenvalue of A, then prove the complex conjugate of lamda is also an eigenvalue of A. In other words, the non real eigenvalues of A occur in conjugate pairs.

The hypothesis $A$ orthogonal is irrelevant, we only need to know $A$ is a real matrix. If $\lambda$ is a complex eigenvalue of $A\in \mathbb{R}^{n\times n}$ then, there exists $0\neq z\in \mathbb{C}^{n\times 1}$ such that $Az=\lambda z$ . Taking conjugate, $A\bar{z}=\bar{\lambda}\bar{z}$ with $\bar{z}\neq 0$ which implies that $\bar{\lambda}$ is an eigenvalue of $A$ .

Fernando Revilla