Orthogonal problem

• Apr 21st 2006, 06:40 AM
dopi
Orthogonal
If A and B are both orthogonal nxn matrices.

Can anyone show That AB is orthogonal...

Thanks
• Apr 21st 2006, 11:57 AM
dopi
Orthogonal problem
If A and B are both orthogonal nxn matrices.

Can anyone show That AB is orthogonal...iv been stuck on this question for a while now

Thanks
• Apr 21st 2006, 12:19 PM
CaptainBlack
Quote:

Originally Posted by dopi
If A and B are both orthogonal nxn matrices.

Can anyone show That AB is orthogonal...iv been stuck on this question for a while now

Thanks

An $\displaystyle n\times n$ $\displaystyle A$ is orthogonal iff

$\displaystyle AA^T=I$

Now suppose $\displaystyle A$ and $\displaystyle B$ be orthogonal, and let
$\displaystyle C=AB$. Now $\displaystyle C^T=B^TA^T$, so:

$\displaystyle CC^T=ABB^TA^T=AIA^T=AA^T=I$

Hence $\displaystyle C$ is orthogonal.

RonL
• Apr 21st 2006, 01:40 PM
dopi
Quote:

Originally Posted by CaptainBlack
An $\displaystyle n\times n$ $\displaystyle A$ is orthogonal iff

$\displaystyle AA^T=I$

Now suppose $\displaystyle A$ and $\displaystyle B$ be orthogonal, and let
$\displaystyle C=AB$. Now $\displaystyle C^T=B^TA^T$, so:

$\displaystyle CC^T=ABB^TA^T=AIA^T=AA^T=I$

Hence $\displaystyle C$ is orthogonal.

RonL

Say if you add A and B (A+B)...would that be Orthognal?
• Apr 21st 2006, 10:07 PM
CaptainBlack
Quote:

Originally Posted by dopi
Say if you add A and B (A+B)...would that be Orthognal?

Not generally

RonL
• Apr 22nd 2006, 02:26 AM
TD!
From the definition, we have for A and B:

$\displaystyle \begin{array}{l} AA^{ - 1} = A^{ - 1} A = I \\ BB^{ - 1} = B^{ - 1} B = I \\ \end{array}$

$\displaystyle \left( {AB} \right)\left( {AB} \right)^{ - 1} = ABB^{ - 1} A^{ - 1} = AIA^{ - 1} = AA^{ - 1} = I$
• Apr 22nd 2006, 03:19 AM
CaptainBlack
Don't make duplicate posts
Don't post the same question in two different fora.

I have merged these two because they both have responses.

RonL
• Apr 22nd 2006, 03:26 AM
CaptainBlack
Quote:

Originally Posted by TD!
From the definition, we have for A and B:

$\displaystyle \begin{array}{l} AA^{ - 1} = A^{ - 1} A = I \\ BB^{ - 1} = B^{ - 1} B = I \\ \end{array}$

$\displaystyle \left( {AB} \right)\left( {AB} \right)^{ - 1} = ABB^{ - 1} A^{ - 1} = AIA^{ - 1} = AA^{ - 1} = I$

This is true of any non-singlar square matrices $\displaystyle A$ and $\displaystyle B$!

RonL