# Prove that every unitary matrix is diagonalisable by a unitary matrix

• Hall
All you had to do was show that every element in ##C C^{*}## is orthogonal to every other element and that the norm of each element is 1.f

#### Hall

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}##.

The question doesn't imply that a unitary matrix is only diagonalisable by a unitary transformation matrix. In fact, you should be able to see why that cannot be the case.

You cannot, therefore, prove that ##C## is unitary. Instead, you must show that you can find a unitary matrix ##C## that does the job.

If ##C = [u_1 ~ \cdots u_n]##, where ##u_i## is normalised eigenvector of A, then we can show that ##C^{-1} A C## will be a diagonal matrix, made up of eigenvalues of A.
But how do we show that CC* = I. We have only two facts with us, that every ##u_i## is orthogonal to every other ##u_j## and norm of each ##u_i## is 1.

So, we have
$$C^{*}= \begin{bmatrix} \overline{u_1} \\ \vdots\\ \overline{u_n}\\ \end{bmatrix}$$

$$CC^{*}= u_1 \overline{u_1} + \cdots u_n \overline{u_n}$$

I really have no idea of what to do next.

If ##C = [u_1 ~ \cdots u_n]##, where ##u_i## is normalised eigenvector of A, then we can show that ##C^{-1} A C## will be a diagonal matrix, made up of eigenvalues of A.
But how do we show that CC* = I. We have only two facts with us, that every ##u_i## is orthogonal to every other ##u_j## and norm of each ##u_i## is 1.

So, we have
$$C^{*}= \begin{bmatrix} \overline{u_1} \\ \vdots\\ \overline{u_n}\\ \end{bmatrix}$$

$$CC^{*}= u_1 \overline{u_1} + \cdots u_n \overline{u_n}$$

I really have no idea of what to do next.
You seem to be confusing matrix multiplication with matrix element multiplication.

What can you say about the eigenvalues of a unitary matrix?

If you can prove that a unitary matrix has a set of mutually orthonormal eigenvectors (or assume this), then you should be able to complete the proof using what you have.

• malawi_glenn
Well, I can write C and C* in expanded form to avoid confusion:
$$C = \begin{bmatrix} u_{11} & u_{12} & \cdots & u_{1n} \\ u_{21} & u_{22} & \cdots & u_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ u_{n1} & u_{n2} & \cdots & \ u_{nn} \\ \end{bmatrix}$$
Where the first column is the first normalised eigenvector, and so on.

$$C^{*} = \begin{bmatrix} \overline{u_{11} } & \overline{ u_{21} } & \cdots & \overline{ u_{n1} } \\ \overline{ u_{12} } & \overline{ u_{22} } & \cdots & \overline{ u_{n2} } \\ \vdots \\ \overline{ u_{1n} } & \overline{ u_ {2n} } & \cdots & \overline{ u_{nn} } \\ \end{bmatrix}$$

The ij th entry of CC* is
$$CC^{*}_{ij} = [ u_{i1} ~u_{i2} \cdots u_{in} ] \times \begin{bmatrix} \overline{ u_{j1} } \\ \overline{ u_{j2} } \\ \vdots \\ \overline { u_{jn} } \\ \end{bmatrix}$$

$$CC^{*}_{ij} = \sum u_{ik}~ \overline{ u_{jk} }$$

Then?

It might be clearer if you looked at ##C^*C##. Also, you should have complex conjugates of the elements in ##C^*##.

Not getting your hint. I have shown my attempts.

Not getting your hint. I have shown my attempts.
You must learn to recognise an inner product when you see one!

• malawi_glenn
Inner product: ##\langle u_i, u_j \rangle = 0##.

It is easy to show ##C^{*} C = I##, that is:
$$C_{\iota}^{*}= \begin{bmatrix} \overline{ u_{1\iota} } & \overline{ u_{2\iota} }& \cdots \overline{n \iota} \end{bmatrix} \times \begin{bmatrix} u_{1j} \\ u_{2j}\\ \vdots \\ u_{nj} \\ \end{bmatrix}= C^j$$

$$(C^{*} C)_{\iota ~j} = u_{1j} \overline{ u_{1\iota}} + u_{2j} \overline{ u_{2\iota}} + \cdots + u_{nj} \overline{ u_ { n \iota} } = u_i \cdot \overline{u_j} = \langle u_i, u_j \rangle = 0$$

Thus, all ij entry of ##C^{*}C## is zero, and by taking ##\iota= j## in the above formula I will get ##\iota- \iota## entry as 1.

So, ## C^{*} C = I##

I don't understand why ##C C^{*} = I## was so hard to prove.