# Thread: Show that A is an orthogonal matrix

1. ## Show that A is an orthogonal matrix

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

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

My working:

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

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

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

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

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

2. You've correctly computed $A^{T}.$ Now form both products $AA^{T}$ and $A^{T}A.$ For the $AA^{T}$ product, note that $(I-B)$ and $(I+B)$ 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 $A^{T}A$ 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 $A^{T}.$ Now form both products $AA^{T}$ and $A^{T}A.$ For the $AA^{T}$ product, note that $(I-B)$ and $(I+B)$ 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 $A^{T}A$ version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...
$(I+B)(I-B)=(I-B)(I+B)$

Taking inverses of both sides,

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

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

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

$=I$

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

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

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

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

In fact, the eigenvalues of a real skew symmetric are of the form $\alpha i\;(\alpha\in\mathbb{R})$ .
Very nice, but isn't the invertibility of $(I-B)$ assumed by the definition of $A$? Maybe you need to prove that $(I+B)$ 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 $(I-B)$ assumed by the definition of $A$?

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

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

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