Prove that every unitary matrix is diagonalisable by a unitary matrix

[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 $C^* =C^{-1}$.

 

[PeroK]

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.

 

[Hall]

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.

 

[PeroK]

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.

 

[Hall]

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?

 

[PeroK]

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

 

[Hall]

Not getting your hint. I have shown my attempts.

 

[PeroK]

Not getting your hint. I have shown my attempts.

You must learn to recognise an inner product when you see one!

 

[Hall]

Inner product: $\langle u_i, u_j \rangle = 0$.

 

[PeroK]

If you are still stuck, take at look at the alternative characterisations of a unitary matrix:

https://en.wikipedia.org/wiki/Unitary_matrix#Equivalent_conditions

 

[Hall]

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.