Linear Algebra orthogonal matrix

[slamminsammya]

Homework Statement

Suppose that A is a real n by n matrix which is orthogonal, symmetric, and positive definite. Prove that A is the identity matrix.

Homework Equations

Orthogonality means $A^t=A^{-1}$, symmetry means $A^t=A$, and positive definiteness means $x^tAx>0$ whenever x is a nonzero vector.

The Attempt at a Solution

Messing around with inner products, trying to show that the matrix $A-I$ is the zero matrix. Help is appreciated.

 

[I like Serena]

Welcome to PF, slamminsammya!

Are you aware of the spectral theorem for symmetric matrices?

What can you say about the eigenvalues of the matrix A?

 

[micromass]

Let's prove this is steps. Can you first show that 1 is the only eigenvalue of A?

Edit: ILS was first

 

[slamminsammya]

Yes, I have been able to show that 1 is the only eigenvalue, but I just can't seem to get from there to showing that A therefore fixes each vector. The proof that 1 is the only eigenvalue goes like this:

Suppose Ax=cx for a scalar c. Then A(Ax)=c^2x=x, so that the only possible eigenvalues are 1 or -1. But -1 cannot be an eigenvalue, since if that were so then <x, Ax>=-|x|^2 < 0, contradicting positive definiteness.

From here, wikipedia tells me that from the spectral theorem for symmetric matrices, A must be similar to the identity matrix (since the eigenvalue 1 has multiplicity n). Does this mean that A must be the identity?

 

[slamminsammya]

Yes it does... Thanks

 

[slamminsammya]

Out of curiosity, it seems like there should be a way of proving this without invoking the spectral theorem for symmetric matrices. Does anyone know a proof?

 

[Maybe_Memorie]

I solved this using the Cayley-Hamilton theorem; every square matrix A satisfies it's characteristic polynomial.

You said A^t = A, because it's symmetric
and you've said A^t = A^-1 for orthogonality.

So combining these results you obtain A = A^-1
and therefore A² = I, where I is the identity matrix.

Your minimal polynomial is therefore m(x) = x² - 1
The roots of this are your eigenvalues, which are 1 and -1, but we reject -1 as it contradicts positive definiteness. So our eigenvalue is 1, which according to our minimal polynomial has multiplicity 1.
So this means the jordan normal form of A is the Identity matrix, as the jordan form is n 1's along the diagonal.

However this only shows that A is similar to I, however, if we are working with the same basis then it surely means A = I...

 

[I like Serena]

I solved this using the Cayley-Hamilton theorem; every square matrix A satisfies it's characteristic polynomial.

Nice!

For the record, at this moment I would not know how to do it without an advanced theorem.

From here, wikipedia tells me that from the spectral theorem for symmetric matrices, A must be similar to the identity matrix (since the eigenvalue 1 has multiplicity n). Does this mean that A must be the identity?

Yes it does... Thanks

However this only shows that A is similar to I, however, if we are working with the same basis then it surely means A = I...

It's still not trivial that similarity to the identity matrix implies identity...

 

[Maybe_Memorie]

Nice!

And to think last week I didn't understand it, now it's my first approach to most of these types of questions.

It's still not trivial that similarity to the identity matrix implies identity...

Right so we need to use the fact that it's similar to the identity to show that it's equal.

This is all I can come up with;

A is similar to I, so A = P^-1IP where P is the transition matrix.
So then, A² = P^-1I²P = P^-1IP = A

But I = A² = P^-1IP = A

Thus, A = I.


Seems correct anyway.

 

[I like Serena]

Yep! Looks correct!

Note:
A is similar to I, so A = P^-1I P where P is the transition matrix.
So A = P^-1I P = P^-1P = I.

 

[slamminsammya]

I solved this using the Cayley-Hamilton theorem; every square matrix A satisfies it's characteristic polynomial.

You said A^t = A, because it's symmetric
and you've said A^t = A^-1 for orthogonality.

So combining these results you obtain A = A^-1
and therefore A² = I, where I is the identity matrix.

Your minimal polynomial is therefore m(x) = x² - 1

I completely agree with your method here, but I don't think that x^2-1 is the characteristic polynomial of A. The characteristic polynomial is the determinant of A-cI, and for any n by n matrix the characteristic polynomial must therefore be of degree n. Forgive me if I am misunderstanding, but I don't quite see how Cayley Hamilton applies in the way you were treating it.

Maybe this is exactly what you were saying, but wouldn't the conclusion be that the characteristic polynomial must be (x-1)^n, and so the only way for A to satisfy this polynomial is if A=I?

 

[micromass]

I completely agree with your method here, but I don't think that x^2-1 is the characteristic polynomial of A. The characteristic polynomial is the determinant of A-cI, and for any n by n matrix the characteristic polynomial must therefore be of degree n. Forgive me if I am misunderstanding, but I don't quite see how Cayley Hamilton applies here.

She said that $x^2-1$ is the minimal polynomial (or correctly: that the minimal polynomial divides $x^2-1$). This is correct.