# A is unitary iff the diagonals are complex unit numbers

• Nov 4th 2011, 01:52 PM
dwsmith
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?
• Nov 4th 2011, 02:00 PM
Ackbeet
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?
• Nov 4th 2011, 02:01 PM
dwsmith
Re: A is unitary iff the diagonals are complex unit numbers
Quote:

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.
• Nov 4th 2011, 02:05 PM
Ackbeet
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?
• Nov 4th 2011, 02:07 PM
dwsmith
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.
• Nov 4th 2011, 02:10 PM
Ackbeet
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?
• Nov 4th 2011, 02:12 PM
dwsmith
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.
• Nov 4th 2011, 02:36 PM
Deveno
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.
• Nov 5th 2011, 09:09 AM
dwsmith
Re: A is unitary iff the diagonals are complex unit numbers
Quote:

Originally Posted by 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?
• Nov 5th 2011, 09:12 AM
Ackbeet
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?
• Nov 5th 2011, 09:23 AM
dwsmith
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
• Nov 5th 2011, 11:00 AM
Ackbeet
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?
• Nov 5th 2011, 11:37 AM
dwsmith
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.
• Nov 5th 2011, 11:54 AM
Ackbeet
Re: A is unitary iff the diagonals are complex unit numbers
Quote:

Originally Posted by 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.