Improper orthogonal matrix plus identity noninvertible?

[ArcanaNoir]

Homework Statement

If P is an orthogonal matrix with detP = -1, show that I+P has no inverse. (Hint: show that (P^t)(I+P)=(I+P)^t)

P^t is P transposed.
I is the identity matrix given by PP^t=I
a^-1 means inverse a
a, b, P and such letters, capital or otherwise, are all matrices, limit to square matrices for our purposes here.

Homework Equations

If I knew all the relevant equations I wouldn't be stuck...

For the transpose, I know (ab)^t=(b^t)(a^t) , (a+b)^t= a^t + b^t , (a^t)^-1 = (a^-1)^t
and A= .5(A + A^t) + .5(A - A^t). (maybe this is the important one?)

I know for orthogonal matrices P^t=P^-1 and P(P^t)=I

For determinates, I know det(I)=1 , det(A^-1)=1/(detA) , det(AB)=det(A)det(B) , det(A^t)=det(A) and the determinate of orthogonal matrices is +1 or -1, called proper when positive and improper when negative.

Also, the rank of an invertible matrix must equal the number of rows (or columns, same thing), and the determinate mustn't be 0.

The Attempt at a Solution

Now, to complicate matters, we have not discussed orthogonal matrices and some relevant related topics in class due to schedule limitations and guest high school students. This question is part of a list of similar questions for a special project only for the undergrads in the class. Everything I know about orthogonal matrices I read out of an assortment of textbooks in the library. I successfully completed other proofs using simply structured equations like the ones above and I would like to solve this problem similarly, leaving out the complicated bits about orthonormal columns and something or others that I haven't quite understood. If it cannot be done simply, then I am willing to delve deeper into that stuff. I do want, however, to avoid geometric arguments and explanations like the plague. I am not a visual thinker, and have zero basis for understanding linear algebra in this way, so to try to do so now would not be the best approach.

So, I'm not looking for someone to just give me the proof, I'm braver than that. I just don't know what I'm not seeing. Surely I'm missing some equation or relationship here. I feel like the determinate is important but I don't know any applicable additive properties of determinates or ranks or anything. Replacing I with PP^t got me no further. It seems interesting that detI = 1 and detP = -1 which when added together = 0 and a determinate of zero gives a noninvertible matrix, but I know you can't just go around adding determinates... Right?

 

[micromass]

Hi ArcanaNoir!

Can you show the hint first?? That is, can you show

$$P^t(I+P)=(I+P)^t$$

You'll only need to use that $P^tP=I$ for this.

 

[ArcanaNoir]

Okay, okay, this is good :D

Can you make sure I didn't violate any properties?

P^t(I+P)=(I+P)^t
P^t(PP^t+P)=(PP^t+P)^t
P^tPP^t+P^tP=(PP^t)^t+P^t
P^tPP^t+P^tP=(P^t)^tP^t+P^t
P^tPP^t+PP^t=PP^t+P^t
P^tPP^t=P^t
IP^t=P^t
P^t=P^t

Yay! I couldn't make that work yesterday. But does the det come into play? I saw a formula today stating det(A+I)= a crazy series. Things about the trace too. I don't think it has to be as bad as all that, but I can't find any reference to the specific cases of orthogonal matrices and any special determinant additive property that may only apply to them. Is that even what I should be looking for?

 

[micromass]

Okay, okay, this is good :D

Can you make sure I didn't violate any properties?

P^t(I+P)=(I+P)^t
P^t(PP^t+P)=(PP^t+P)^t
P^tPP^t+P^tP=(PP^t)^t+P^t
P^tPP^t+P^tP=(P^t)^tP^t+P^t
P^tPP^t+PP^t=PP^t+P^t
P^tPP^t=P^t
IP^t=P^t
P^t=P^t

Uh, ok, this is correct. But you're making it much too difficult!

$$P^t(I+P)=P^t+P^tP=P^t+I=(P+I)^t=(I+P)^t$$

Yay! I couldn't make that work yesterday. But does the det come into play? I saw a formula today stating det(A+I)= a crazy series. Things about the trace too. I don't think it has to be as bad as all that, but I can't find any reference to the specific cases of orthogonal matrices and any special determinant additive property that may only apply to them. Is that even what I should be looking for?

No, you don't need to do crazy things with series and the trace. Just calculate the determinant of both sides of

$$P^t(I+P)=(I+P)^t$$

and remember that $det(A)=det(A^t)$.

 

[ArcanaNoir]

So...like... if
det(a)=det(a^t)
then det((I+P))=det((I+P)^t)

And since (I+P)^t=...
I keep trying to say det(a+b)=det(a)+det(b) and I know that's not allowed. How do I get around that?

 

[micromass]

You do know that det(AB)=det(A)det(B).

What do you get if you take the determinant of both sides of

$$P^t(I+P)=(I+P)^t$$

 

[ArcanaNoir]

Ooo, I get detP=1 !
Since we stated detP=-1, using definitions of inverse and such I can prove it by contradiction, right? Thanks!

 

[micromass]

Yes, that's good!

 

[ArcanaNoir]

I'm going to blame that headache on my prof. He said these were all direct proofs. That had me thinking I had to arrive directly at a noninvertible quantity, like det = 0 or something. Pffff. :)

 

[delinne]

I understand how you get det(P)=1. But which definition do you have to use to prove that P+I is invertible when det(P) is 1?

And the hint that is given is this just improvisation or is it also based om some rule?

 

[AlephZero]

I'm going to blame that headache on my prof. He said these were all direct proofs.

It is a direct proof. You have det(P^T) det(I+P) = det(I+P)^T

You know det(P) = det(P^t) = -1 and det(I+P) = det(I+P)^T

So -det(I+P) = det(I+P)
det(I+P) = 0

 

[ArcanaNoir]

Yes, it was a direct proof. My final solution was somewhat different than what I thought when I wrote here.