Eigenvalues of an orthogonal matrix

[etotheipi]

Homework Statement

Show that the three eigenvalues of a real orthogonal 3x3 matrix are $e^{i\alpha}$, $e^{-i\alpha}$, and $+1$ or $-1$, where $\alpha \in \mathbb{R}$.

Relevant Equations

N/A

I'm fairly stuck, I can't figure out how to start. I called the matrix $\mathbf{A}$ so then it gives us that $\mathbf{A}\mathbf{A}^\intercal = \mathbf{I}$ from the orthogonal bit. I tried 'determining' both sides... $$(\det(\mathbf{A}))^{2} = 1 \implies \det{\mathbf{A}} = \pm 1$$I don't think this helps me, since I need to show that $\det{(\mathbf{A} - \lambda \mathbf{I})} = 0$ for the values of $\lambda$ in the question and as far as I'm aware there is no rule for $\det{(\mathbf{a}+\mathbf{b})}$.

I don't know if there is a general form of an orthogonal matrix I can use as a way in (I suspect there isn't, though this might be wrong!). I wondered whether anyone could give me a hint! Thank you.

 

[PeroK]

Homework Statement:: Show that the three eigenvalues of a real orthogonal 3x3 matrix are $e^{i\alpha}$, $e^{-i\alpha}$, and $+1$ or $-1$, where $\alpha \in \mathbb{R}$.
Relevant Equations:: N/A

I'm fairly stuck, I can't figure out how to start. I called the matrix $\mathbf{A}$ so then it gives us that $\mathbf{A}\mathbf{A}^\intercal = \mathbf{I}$ from the orthogonal bit. I tried 'determining' both sides... $$(\det(\mathbf{A}))^{2} = 1 \implies \det{\mathbf{A}} = \pm 1$$I don't think this helps me, since I need to show that $\det{(\mathbf{A} - \lambda \mathbf{I})} = 0$ for the values of $\lambda$ in the question and as far as I'm aware there is no rule for $\det{(\mathbf{a}+\mathbf{b})}$.

I don't know if there is a general form of an orthogonal matrix I can use as a way in (I suspect there isn't, though this might be wrong!). I wondered whether anyone could give me a hint! Thank you.

Hint: what sort of characteristic equation do you get for a 3x3 matrix?

 

[etotheipi]

Hint: what sort of characteristic equation do you get for a 3x3 matrix?

I get

$$\det \left( \begin{pmatrix}a & b & c\\d & e & f\\g & h & i\\\end{pmatrix} - \begin{pmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\\\end{pmatrix} \right) = 0 $$ $$= (a-\lambda)[(e-\lambda)(i-\lambda) - hf] - b[d(i-\lambda) - gf] + c(dh - g(e-\lambda)) = 0$$

 

[PeroK]

I get

$$\det \left( \begin{pmatrix}a & b & c\\d & e & f\\g & h & i\\\end{pmatrix} - \begin{pmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\\\end{pmatrix} \right) = 0 $$ $$= (a-\lambda)[(e-\lambda)(i-\lambda) - hf] - b[d(i-\lambda) - gf] + c(dh - g(e-\lambda)) = 0$$

How would you, in general, describe any such equation?

 

[etotheipi]

How would you, in general, describe any such equation?

I expanded it all out (hope I didn't mess anything up!) and got

$\lambda^{3} -(a+e+i)\lambda^{2} + (ai + ea + ei - hf - bd - gc)\lambda + (aei + gfb + cdh - hfa - bdi - cga) = 0$
The coefficient of $\lambda^{2}$ I can tell to be the trace of $\mathbf{A}$, the final constant term is I think the determinant, whilst the coefficient of $\lambda$ I can't assign to anything of meaning.

 

[PeroK]

I expanded it all out (hope I didn't mess anything up!) and got

$\lambda^{3} -(a+e+i)\lambda^{2} + (ai + ea + ei - hf - bd - gc)\lambda + (aei + gfb + cdh - hfa - bdi - cga) = 0$
The coefficient of $\lambda^{2}$ I can tell to be the trace of $\mathbf{A}$, the final constant term is I think the determinant, whilst the coefficient of $\lambda$ I can't assign to anything of meaning.

It's a cubic equation!

 

[etotheipi]

It's a cubic equation!

Right so then we have 2 complex roots and 1 real root. The coefficients are real so that fixes $e^{i\alpha}$ and $e^{-i\alpha}$ to be two arbitrary solutions, so all that remains is to show that $\pm 1$ is the other solution. And I think we don't need to prove that algebraically, a geometrical argument should do?

 

[PeroK]

Right so then we have 2 complex roots and 1 real root. The coefficients are real so that fixes $e^{i\alpha}$ and $e^{-i\alpha}$ to be two arbitrary solutions, so all that remains is to show that $\pm 1$ is the other solution. And I think we don't need to prove that algebraically, a geometrical argument should do?

Not quite. It means you have one real root and a pair of conjugate roots.

Do you know the properties of an orthogonal matrix? There's a key (alternative defining) property.

 

[etotheipi]

Not quite. It means you have one real root and a pair of conjugate roots.

Do you know the properties of an orthogonal matrix? There's a key (alternative defining) property.

Yeah, I just realized I forgot about the modulus, so at the moment we're stuck with a real solution an two complex ones of the form $\beta e^{i \alpha}$ and $\beta e^{-i \alpha}$.

I went onto Wikipedia and found that for a vector $\vec{v}$ and its transpose $\vec{v}^\intercal$, if $Q$ is orthogonal then

$\vec{v}^\intercal \vec{v} = \vec{v}^\intercal Q^\intercal Q \vec{v}$

I'll play around with that for a little bit and see if I can get anything useful, assuming that's the one you were making reference to!

 

[PeroK]

Yeah, I just realized I forgot about the modulus, so at the moment we're stuck with a real solution an two complex ones of the form $\beta e^{i \alpha}$.

I went onto Wikipedia and found that for a vector $\vec{v}$ and its transpose $\vec{v}^\intercal$, if $Q$ is orthogonal then

$\vec{v}^\intercal \vec{v} = \vec{v}^\intercal Q^\intercal Q \vec{v}$

I'll play around with that for a little bit and see if I can get anything useful, assuming that's the one you were making reference to!

Yes, although I would think of that it terms of the inner product (and the the norm).

 

[etotheipi]

Yes, although I would think of that it terms of the inner product (and the the norm).

Yeah that makes more sense, because then $||A\vec{v}||^{2} = ||\vec{v}||^{2}$ since the $A$ and its transpose do some cancelling if $A$ is orthogonal.

In that case, can't we just say the square of the norm is going to just be $\lambda^{2} ||\vec{v}||^{2}$, which is just $\lambda^{2}$ times itself so the modulus of lambda has to be 1, and this gives us all of our results?

 

[PeroK]

Yeah that makes more sense, because then $||A\vec{v}||^{2} = ||\vec{v}||^{2}$ since the $A$ and its transpose do some cancelling if $A$ is orthogonal.

In that case, can't we just say the square of the norm is going to just be $\lambda^{2} ||\vec{v}||^{2}$, which is just $\lambda^{2}$ times itself so the modulus of lambda has to be 1, and this gives us all of our results?

Yes need to be careful because $\lambda$ may be complex. The simplest approach is to consider $A$ as a unitary matrix with real coefficients.

 

[etotheipi]

Yes need to be careful because $\lambda$ may be complex. The simplest approach is to consider $A$ as a unitary matrix with real coefficients.

Yes, completely forgot about that part! I should really write $||\vec{v}||^2 = ||A \vec{v}||^2 = ||\lambda \vec{v}||^2 = \lambda \lambda^* ||\vec{v}||^2 = |\lambda|^2 ||\vec{v}||^2$. Thanks for the help!

 

[vela]

Right so then we have 2 complex roots and 1 real root.

How did you conclude that two of the roots are complex? A cubic could have three real roots.

 

[etotheipi]

How did you conclude that two of the roots are complex? A cubic could have three real roots.

I suppose you're right, I just assumed from how the question was phrased that there were only two possibilities for real roots ($1$ and $-1$), and this indeed turned out to be the case since there are only two re`als with modulus 1.

But the point stands, I should have been more careful at that stage .

 

[PeroK]

I suppose you're right, I just assumed from how the question was phrased that there were only two possibilities for real roots ($1$ and $-1$), and this indeed turned out to be the case since there are only two re`als with modulus 1.

But the point stands, I should have been more careful at that stage .

There are the degenerate cases where $\lambda = \pm 1$ only. Although these do meet the criteria in each case.

 

[etotheipi]

There are the degenerate cases where $\lambda = \pm 1$ only.

Oh, do you mean if we got something like $\alpha = \pi$ or $\alpha = 0$ so that we would indeed have three real roots, which might be $(\lambda_1, \lambda_2, \lambda_3) = (1,1,1), (1,1,-1), (1,-1,-1), (-1,-1,-1)$. Hadn't thought of that either!

 

[PeroK]

Oh, do you mean if we got something like $\alpha = \pi$ or $\alpha = 0$ so that we would indeed have three real roots, which might be $(\lambda_1, \lambda_2, \lambda_3) = (1,1,1), (1,1,-1), (1,-1,-1), (-1,-1,-1)$. Hadn't thought of that either!

I didn't think of the degenerate cases, but yes you can make it work like that for those as well.

 

[StoneTemplePython]

There are the degenerate cases where $\lambda = \pm 1$ only. Although these do meet the criteria in each case.

These occur iff the real orthogonal matrix is symmetric. I don't really view involutions as "degenerate" though. In fact involutions are quite nice.