# Thread: Show that A is an orthogonal matrix

1. ## Show that A is an orthogonal matrix

$\displaystyle A=(I-B)^{-1}(I+B)$

Show that A is always an orthogonal matrix (ie $\displaystyle A^{T}A=I$) if $\displaystyle B^{T}=-B$.

My working:

$\displaystyle A^{T}=(I+B)^{T}[(I-B)^{-1}]^{T}$

$\displaystyle =(I+B)^{T}[(I-B)^{T}]^{-1}$

$\displaystyle =(I^T+B^T)(I^T-B^T)^{-1}$

$\displaystyle =(I-B)(I+B)^{-1}$

I don't seem to be getting anywhere. Any suggestions?

2. You've correctly computed $\displaystyle A^{T}.$ Now form both products $\displaystyle AA^{T}$ and $\displaystyle A^{T}A.$ For the $\displaystyle AA^{T}$ product, note that $\displaystyle (I-B)$ and $\displaystyle (I+B)$ commute (just multiply it out both ways, and you'll see why). That will get you the result you want for that product. As for the $\displaystyle A^{T}A$ version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...

3. Originally Posted by Ackbeet
You've correctly computed $\displaystyle A^{T}.$ Now form both products $\displaystyle AA^{T}$ and $\displaystyle A^{T}A.$ For the $\displaystyle AA^{T}$ product, note that $\displaystyle (I-B)$ and $\displaystyle (I+B)$ commute (just multiply it out both ways, and you'll see why). That will get you the result you want for that product. As for the $\displaystyle A^{T}A$ version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...
$\displaystyle (I+B)(I-B)=(I-B)(I+B)$

Taking inverses of both sides,

$\displaystyle (I-B)^{-1}(I+B)^{-1}=(I+B)^{-1}(I-B)^{-1}$ ---------- (1)

Thus, $\displaystyle A^{T}A=(I-B)(I+B)^{-1}(I-B)^{-1}(I+B)$

$\displaystyle =(I-B)(I-B)^{-1}(I+B)^{-1}(I+B)$ [Using (1)]

$\displaystyle =I$

4. Very nice! I think you're done.

5. To complete the proof, it is necessary to say that $\displaystyle (I-B)^{-1}$ always exists if $\displaystyle B^t=-B$ . Equivalently, $\displaystyle \det (I-B)\neq 0$ or $\displaystyle \lambda=1$ is not an eigenvalue of $\displaystyle B$.

In fact, the eigenvalues of a real skew symmetric are of the form $\displaystyle \alpha i\;(\alpha\in\mathbb{R})$ .

6. Originally Posted by FernandoRevilla
To complete the proof, it is necessary to say that $\displaystyle (I-B)^{-1}$ always exists if $\displaystyle B^t=-B$ . Equivalently, $\displaystyle \det (I-B)\neq 0$ or $\displaystyle \lambda=1$ is not an eigenvalue of $\displaystyle B$.

In fact, the eigenvalues of a real skew symmetric are of the form $\displaystyle \alpha i\;(\alpha\in\mathbb{R})$ .
Very nice, but isn't the invertibility of $\displaystyle (I-B)$ assumed by the definition of $\displaystyle A$? Maybe you need to prove that $\displaystyle (I+B)$ is invertible. Certainly your proof would be sufficient to show that as well, I think.

7. Originally Posted by Ackbeet
Very nice, but isn't the invertibility of $\displaystyle (I-B)$ assumed by the definition of $\displaystyle A$?

Well, perhaps, but I don't see it very clear.

Maybe you need to prove that $\displaystyle (I+B)$ is invertible. Certainly your proof would be sufficient to show that as well, I think.

Right, in the same way: $\displaystyle -1\neq \alpha i$ for all $\displaystyle \alpha \in \mathbb{R}$ .