1. ## Orthogonal

If A and B are both orthogonal nxn matrices.

Can anyone show That AB is orthogonal...

Thanks

2. ## 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

3. 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

4. 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?

5. Originally Posted by dopi
Say if you add A and B (A+B)...would that be Orthognal?
Not generally

RonL

6. From the definition, we have for A and B:

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

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

7. ## 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

8. 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