Proof of the properties of conjugate matrices

[diogomgf]

Hello math helpers,

I'm having problems with demonstrating the following property:
[MATH] \bar{AB} = \bar{A} \bar {B} [/MATH]
Where AB stands for the product of the matrices A and B.

Could any one help me with this?
Thanks

 

[pka]

I'm having problems with demonstrating the following property:
[MATH] \bar{AB} = \bar{A} \bar {B} [/MATH]Where AB stands for the product of the matrices A and B.

A note on notation: Use [ tex]\overline{A\cdot B}[/tex] for \(\displaystyle \overline{A\cdot B}\)
To prove the statement we need that the conjugate of a sum or product is the sum or product of conjugates:
\(\displaystyle \overline{z+ w}=\overline{z}+\overline{ w}\) and \(\displaystyle \overline{z\cdot w}=\overline{z}\cdot\overline{ w}\)

 

[diogomgf]

A note on notation: Use [ tex]\overline{A\cdot B}[/tex] for \(\displaystyle \overline{A\cdot B}\)

I'm struggling to find the correct way to go about notations. I'm new to online forums in general, and what I got came from a LaTeX symbol document, where I searched for the conjugate of a number and got the \bar notation.

To prove the statement we need that the conjugate of a sum or product is the sum or product of conjugates:
\(\displaystyle \overline{z+ w}=\overline{z}+\overline{ w}\) and \(\displaystyle \overline{z\cdot w}=\overline{z}\cdot\overline{ w}\)

This is what I'm trying to prove, just don't know where to start... I really struggle with finding the correct approach for this kind of problems.

 

[topsquark]

If the worst comes to worst let w = a + ib and z = c + id and work out both sides.

-Dan

 

[pka]

I'm struggling to find the correct way to go about notations. I'm new to online forums in general, and what I got came from a LaTeX symbol document, where I searched for the conjugate of a number and got the \bar notation

I will do the sum for you.
\(\displaystyle \begin{align*}\overline{(a+bi)+(c+di)}&=\overline{(a+c)+i(b+d)} \\&=(a+c)-i(b+d)\\&=(a-bi)+(c-di)\\&=\overline{(a+bi)}+\overline{(c+di)} \end{align*}\)

 

[diogomgf]

If the worst comes to worst let w = a + ib and z = c + id and work out both sides.

-Dan

That part is what I always do, but that doesn't work as a formal proof, unfortunately

 

[topsquark]

That part is what I always do, but that doesn't work as a formal proof, unfortunately

No, it's a formal proof. It's just not "elegant."

-Dan

 

[diogomgf]

No, it's a formal proof. It's just not "elegant."

-Dan

Thanks alot!

 

[diogomgf]

@topsquark

Is there any way to show \(\displaystyle \overline{A\cdot B}\) = \(\displaystyle \overline{A} \cdot \overline{B} \) using the induction method?
And if yes, could you show it?

-Diogo

 

[pka]

Is there any way to show \(\displaystyle \overline{A\cdot B}\) = \(\displaystyle \overline{A} \cdot \overline{B} \) using the induction method?
And if yes, could you show it?

Why do that? Why are you making such a to due out of the operation of conjugates.
Every complex number has a real part & an imaginary part. In fact one development of the complex field defines is as a set of ordered pairs. The first member is the real part and the second is the imaginary part. If \(\displaystyle z=a+bi\) then \(\displaystyle \Re(z)=a~\&~\Im(z)=b\).
Now the conjugate operation simply changes the sign of the imaginary part \(\displaystyle \overline{z}=a-bi\)
If \(\displaystyle z=a+bi~\&~w=c+di\) the \(\displaystyle z\cdot w=(ac-bd)+i(ad+bc)\)
\(\displaystyle \begin{align*}\overline{z\cdot w}&=(ac-bd){\large\bf{-}}i(ad+bc) \\&=(ac-bd)+(-adi-bci)\\&=(a-bi)\cdot(c-di)\\&=\overline{z}\cdot\overline{w} \end{align*}\)

 

[Dr.Peterson]

@topsquark

Is there any way to show \(\displaystyle \overline{A\cdot B}\) = \(\displaystyle \overline{A} \cdot \overline{B} \) using the induction method?
And if yes, could you show it?

-Diogo

I notice that no one has explicitly dealt with the fact that your A and B are matrices, not just numbers. Is that what you are concerned about, thinking that more has to be said, perhaps using induction on the size of the matrices?

I don't think induction is needed or helpful, but you do need to deal with the matrices. Just use the definition of matrix multiplication, together with the facts that, for complex numbers, [MATH]\overline{a\cdot b} = \overline{a}\cdot\overline{b}[/MATH] and [MATH]\overline{a+b} = \overline{a}+\overline{b}[/MATH], to show that the same property extends to matrices.

What is that definition?

 

[diogomgf]

@Dr.Peterson
Yes, my main problem is indeed the fact that A and B are matrices, and how to present the formal proof...

(AB)_i,j = [MATH] \sum_{k=1}^n [/MATH] ( a_i,k . b_k,j ) is the definition of matrix multiplication...

 

[pka]

@Dr.Peterson
Yes, my main problem is indeed the fact that A and B are matrices, and how to present the formal proof...
(AB)_i,j = [MATH] \sum_{k=1}^n [/MATH] ( a_i,k . b_k,j ) is the definition of matrix multiplication...

Are you making much to much out of the properties of the conjugate operator.
\(\displaystyle \overline{z+w}=\overline{z}+\overline{w}\) conjugate of a sum equals the sum of the conjugates.
\(\displaystyle \overline{z\cdot w}=\overline{z}\cdot\overline{w}\) conjugate of a product equals the product of the conjugates.
So \(\displaystyle \overline {\sum\limits_{k = 1}^N {{z_k}} } = \sum\limits_{k = 1}^N {\overline {{z_k}} }\)

If \(\displaystyle A=z_{k,j}\) the the conjugate is \(\displaystyle \overline{A~z_{k,j}}=A~\overline{z_{k,j}}\) so just apply the sum & product rules.

 

[Dr.Peterson]

@Dr.Peterson
Yes, my main problem is indeed the fact that A and B are matrices, and how to present the formal proof...

(AB)_i,j = [MATH] \sum_{k=1}^n [/MATH] ( a_i,k . b_k,j ) is the definition of matrix multiplication...

As pka says, just take it one step at a time, applying properties of complex numbers: [MATH]\overline{(AB)_{i,j}} = \overline{\sum_{k=1}^n a_{i,j}\cdot b_{j,k}} = \sum_{k=1}^n \overline{a_{i,j}\cdot b_{j,k}} = \dots[/MATH]
That is perfectly formal, as long as you can justify each step.

 

[diogomgf]

@pka @Dr.Peterson

Thanks for the clear explanation.