# Orthogonal problem

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

Can anyone show That AB is orthogonal...

Thanks
• April 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
• April 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 $n\times n$ $A$ is orthogonal iff

$AA^T=I$

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

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

Hence $C$ is orthogonal.

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

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

$AA^T=I$

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

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

Hence $C$ is orthogonal.

RonL

Say if you add A and B (A+B)...would that be Orthognal?
• April 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
• April 22nd 2006, 02:26 AM
TD!
From the definition, we have for A and B:

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

$
\left( {AB} \right)\left( {AB} \right)^{ - 1} = ABB^{ - 1} A^{ - 1} = AIA^{ - 1} = AA^{ - 1} = I
$
• April 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
• April 22nd 2006, 03:26 AM
CaptainBlack
Quote:

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

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

$
\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 $A$ and $B$!

RonL