A is unitary iff the diagonals are complex unit numbers

**dwsmith** · November 4th 2011, 01:52 PM

Let A be a diagonal matrix.

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

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

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

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

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

Now what can I do?

**Ackbeet** · November 4th 2011, 02:00 PM

Well, if you take

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

what does that tell you?

**dwsmith** · November 4th 2011, 02:01 PM

Originally Posted by Ackbeet

Well, if you take

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

what does that tell you?

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

**Ackbeet** · November 4th 2011, 02:05 PM

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

**dwsmith** · November 4th 2011, 02:07 PM

Originally Posted by Ackbeet

The LHS is what you get from taking the $_{jj}$ component of the product $A\overline{A}^{t}.$ 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.

**Ackbeet** · November 4th 2011, 02:10 PM

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

**dwsmith** · November 4th 2011, 02:12 PM

Originally Posted by Ackbeet

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

Unfortunately no.

**Deveno** · November 4th 2011, 02:36 PM

if $\overline{A}^{t}A=A\overline{A}^{t}=I.$ , 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.

**dwsmith** · November 5th 2011, 09:09 AM

Originally Posted by Deveno

if $\overline{A}^{t}A=A\overline{A}^{t}=I.$ , 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?

**Ackbeet** · November 5th 2011, 09:12 AM

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?

**dwsmith** · November 5th 2011, 09:23 AM

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

**Ackbeet** · November 5th 2011, 11:00 AM

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

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

is

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

because

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

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

**dwsmith** · November 5th 2011, 11:37 AM

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

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

is

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

because

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

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.

**Ackbeet** · November 5th 2011, 11:54 AM

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

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

LinkBack

Thread Tools

Display

A is unitary iff the diagonals are complex unit numbers

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

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

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

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

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

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

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

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

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

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

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

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

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

Similar Math Help Forum Discussions

Proof of Eigenvalues of unitary matrix lie on unit circle

Complex Numbers & Unit Circle

complex conjugates and unitary

Roots of unit complex numbers

Complex Numbers, Unit Circle

Search Tags