A is unitary iff the diagonals are complex unit numbers

Let A be a diagonal matrix.

Prove that A is unitary iff its diagonal entries are complex unit numbers.

Unitary is $\displaystyle \bar{A}^t=A^{-1}$

$\displaystyle \Rightarrow$

Let $\displaystyle a_{jj}=x_{jj}+y_{jj}i, \ \ x_{jj},y_{jj}\in\mathbb{R}$

$\displaystyle \bar{A}^t=\begin{bmatrix}x_{11}-y_{11}i &0&\cdots &0\\0&\ddots & 0&\vdots \\ \vdots & &\ddots & \\ \dots& & & x_{nn}-y_{nn}i\end{bmatrix}$

$\displaystyle \displaystyle A^{-1}=\frac{1}{\prod_{j=1}^n (x_{jj}+y_{jj}i)}\begin{bmatrix}x_{11}+y_{11}i &0&\cdots &0\\0&\ddots & 0&\vdots \\ \vdots & &\ddots & \\ \dots& & & x_{nn}+y_{nn}i\end{bmatrix}$

Now what can I do?

Re: A is unitary iff the diagonals are complex unit numbers

Well, if you take

$\displaystyle a_{jj}\,\overline{a_{jj}}=|a_{jj}|^{2},$

what does that tell you?

Re: A is unitary iff the diagonals are complex unit numbers

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Originally Posted by

**Ackbeet** Well, if you take

$\displaystyle a_{jj}\,\overline{a_{jj}}=|a_{jj}|^{2},$

what does that tell you?

I am not sure what you want me to gleam from that.

Re: A is unitary iff the diagonals are complex unit numbers

The LHS is what you get from taking the $\displaystyle _{jj}$ component of the product $\displaystyle A\overline{A}^{t}.$ What do you know about the RHS?

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**Ackbeet** The LHS is what you get from taking the $\displaystyle _{jj}$ component of the product $\displaystyle A\overline{A}^{t}.$ What do you know about the RHS?

Why would I want to multiple by A? Don't I need to show the adjoint is the inverse?

The RHS will be I if multiplied by A.

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**dwsmith** Why would I want to multiple by A? Don't I need to show the adjoint is the inverse?

The RHS will be I if multiplied by A.

I would agree. If $\displaystyle A$ is just any ol' square matrix, then $\displaystyle B=A^{-1}$ iff $\displaystyle AB=BA=I.$ So, $\displaystyle \overline{A}^{t}=A^{-1}$ iff $\displaystyle \overline{A}^{t}A=A\overline{A}^{t}=I.$ Now do you see where I'm going with this?

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**Ackbeet** I would agree. If $\displaystyle A$ is just any ol' square matrix, then $\displaystyle B=A^{-1}$ iff $\displaystyle AB=BA=I.$ So, $\displaystyle \overline{A}^{t}=A^{-1}$ iff $\displaystyle \overline{A}^{t}A=A\overline{A}^{t}=I.$ Now do you see where I'm going with this?

Unfortunately no.

Re: A is unitary iff the diagonals are complex unit numbers

if $\displaystyle \overline{A}^{t}A=A\overline{A}^{t}=I.$ , what are the diagonal elements of the two matrices on the left, and the matrix on the right?

the equality sign says they are the same matrices, and matrix equality means every entry is equal.

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

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**Deveno** if $\displaystyle \overline{A}^{t}A=A\overline{A}^{t}=I.$ , what are the diagonal elements of the two matrices on the left, and the matrix on the right?

the equality sign says they are the same matrices, and matrix equality means every entry is equal.

I need to show A inverse = the conjugate transpose.

How do you ascertain that from what either of you wrote?

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**dwsmith** I need to show A inverse = the conjugate transpose.

How do you ascertain that from what either of you wrote?

What is the definition of a matrix inverse? How does your book write that?

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**Ackbeet** What is the definition of a matrix inverse? How does your book write that?

No book for this class. I know it as 1/det(A)*A

Re: A is unitary iff the diagonals are complex unit numbers

So far as I know, the definition of a matrix inverse that is typically used is the one I gave in Post # 6. The inverse of 5 is that number such that, when you multiply it by 5, you get 1. So that'd be 1/5, because 5(1/5)=(1/5)5=1. Similarly, the inverse of

$\displaystyle \begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}$

is

$\displaystyle \begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix},$

because

$\displaystyle \begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}\begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix}=\begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix}\begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}=\begin{bmatrix}1 &0\\ 0 &1\end{bmatrix}=I.$

I'm not sure I understand your definition of the matrix inverse. What does the asterisk mean in your definition?

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

Originally Posted by

**Ackbeet** So far as I know, the definition of a matrix inverse that is typically used is the one I gave in Post # 6. The inverse of 5 is that number such that, when you multiply it by 5, you get 1. So that'd be 1/5, because 5(1/5)=(1/5)5=1. Similarly, the inverse of

$\displaystyle \begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}$

is

$\displaystyle \begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix},$

because

$\displaystyle \begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}\begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix}=\begin{bmatrix}1/2 &0\\ 0 &1/3\end{bmatrix}\begin{bmatrix}2 &0\\ 0 &3\end{bmatrix}=\begin{bmatrix}1 &0\\ 0 &1\end{bmatrix}=I.$

I'm not sure I understand your definition of the matrix inverse. What does the asterisk mean in your definition?

1/det(A) is a scalar and A is a matrix and you don't know what I mean by *???? It has to be scalar multiplication.

Re: A is unitary iff the diagonals are complex unit numbers

Quote:

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**dwsmith** 1/det(A) is a scalar and A is a matrix and you don't know what I mean by *???? It has to be scalar multiplication.

Well, ok, but $\displaystyle (1/\det(A))A$ is not the inverse of $\displaystyle A.$ If you look here, you will see the formula $\displaystyle A^{-1}_{ij}=C_{ji}/\det(A),$ where $\displaystyle C_{ij}$ is the cofactor matrix.