- #1

Hall

- 351

- 88

- Homework Statement
- The same as the title.

- Relevant Equations
- A matrix is said to be unitary if A*A = AA* = I. A

A matrix ##A## is diagonalisable if there is a matrix C such that

##\Lambda = C^{-1} A C ##.

Let's assume that ##A## is unitary and diagonalisable, so, we have

## \Lambda = C^{-1} A C ##

Since, ##\Lambda## is made up of eigenvalues of ##A##, which is unitary, we have ## \Lambda \Lambda^* = \Lambda \bar{\Lambda} = I##.

I tried using some, petty, algebra to prove that ##C C* = I## but was unsuccessful. Here is my attempt:

## \Lambda = C^{-1} A C ##

## \Lambda^* = C^* A^* (C^*)^{-1} ##

## \Lambda \Lambda^* = C ^{-1} A C ~ C^* A^* (C^{-1})^{*} ##

## I = C ^{-1} A C ~ C^* A^* (C^{-1})^{*} ##

## I = C^{-1} AC ~ [ C^{-1} A C ]^{*} ##

That implies,

## [ C^{-1} A C ]^{*} = [C^{-1} A C]^{-1} ##

## C^* A^* (C^{-1})^{*} = C^{-1} A^* C##, since ##A^* = A^{-1}##.

But I cannot conclude ##C^* =C^{-1}##.

## \Lambda = C^{-1} A C ##

Since, ##\Lambda## is made up of eigenvalues of ##A##, which is unitary, we have ## \Lambda \Lambda^* = \Lambda \bar{\Lambda} = I##.

I tried using some, petty, algebra to prove that ##C C* = I## but was unsuccessful. Here is my attempt:

## \Lambda = C^{-1} A C ##

## \Lambda^* = C^* A^* (C^*)^{-1} ##

## \Lambda \Lambda^* = C ^{-1} A C ~ C^* A^* (C^{-1})^{*} ##

## I = C ^{-1} A C ~ C^* A^* (C^{-1})^{*} ##

## I = C^{-1} AC ~ [ C^{-1} A C ]^{*} ##

That implies,

## [ C^{-1} A C ]^{*} = [C^{-1} A C]^{-1} ##

## C^* A^* (C^{-1})^{*} = C^{-1} A^* C##, since ##A^* = A^{-1}##.

But I cannot conclude ##C^* =C^{-1}##.