Constructing an Eigenvector of S with Eigenvalue λ1

[Passers_by]

Homework Statement

In diagonalizing a symmetric tensor S, we find that two of the eigenvalues(λ1, and λ2) are equal but the third ( λ3 ) is different. Show that any vector which is normal to n3 is then an eigenvector of S with eigenvalue equal to λ1.

Relevant Equations

S n1=λ1 n1

There is a eigenvector n3 of S with eigenvalue equal to λ3 and a eigenvector n1 of S with eigenvalue equal to λ1. n1 and n3 are orthogonal to each other . Construct the vector v2 so that they're orthogonal to each other(n1,v2 and n3).We can prove that v2 is an eigenvector of S . But how do we prove that it corresponds to the eigenvalue λ1(λ1=λ2)?

 

[Gaussian97]

Well, if $\lambda_3$ has algebraic multiplicity equal to 1, can you have two linearly-independent vectors that are both eigenvectors of $\lambda_3$?

 

[Passers_by]

Well, if $\lambda_1$ has algebraic multiplicity equal to 1, can you have two linearly-independent vectors that are both eigenvectors of $\lambda_1$?

First of all, thank you. There is no mention of algebraic multiplicity in that book. Maybe there's a better way to prove it.

 

[Gaussian97]

I've made a typo, of course the eigenvalue with multiplicity 1 is $\lambda_3$, not $\lambda_1$.

With this in mind, they tell you that $\lambda_1=\lambda_2\neq \lambda_3$, so they are telling you that $\lambda_3$ has multiplicity 1 (there is no other eigenvalue equal) if they use this name or not is not relevant.
I don't know how else one could prove that $v_2$ has eigenvalue $\lambda_1$.

 

[Passers_by]

I've made a typo, of course the eigenvalue with multiplicity 1 is $\lambda_3$, not $\lambda_1$.

With this in mind, they tell you that $\lambda_1=\lambda_2\neq \lambda_3$, so they are telling you that $\lambda_3$ has multiplicity 1 (there is no other eigenvalue equal) if they use this name or not is not relevant.
I don't know how else one could prove that $v_2$ has eigenvalue $\lambda_1$.

I really mean the proof of the whole problem not that the eigenvalue of V2 is λ1. There should be a better way to do this.

 

[Gaussian97]

Have you tried to expand $v$ in an orthogonal base formed by eigenvectors? (since for symmetric matrices, such a base always exist)

 

[Passers_by]

Have you tried to expand $v$ in an orthogonal base formed by eigenvectors? (since for symmetric matrices, such a base always exist)

I don't think there's much difference between what you're saying and the method I use

 

[Gaussian97]

Sometimes in not about how many differences there are, but how relevant those differences are.
Anyway, it's an idea: You can try it and let's see what you can do with it.

 

[romsofia]

I don't think you can prove this one without the following information, which you can try proving as well:
Given a symmetric tensor A, then we know that it has 3 real roots of the characteristic equation, thus we have 3 eigenvalues. Then, we can prove that all 3 eigenvectors that come from two distinct eigenvalues are orthogonal. From here, we may also state that our tensor must have at least 3 eigenvectors that are mutually orthogonal. However, your problem doesn't need you to prove these statements, so maybe use them and move on.

 

[vela]

I really mean the proof of the whole problem not that the eigenvalue of V2 is λ1. There should be a better way to do this.

What do you mean by "proof of the whole problem"? I'm not sure what you mean by that. Can you tell us what you want to prove as mathematical statements?

 

[Passers_by]

What do you mean by "proof of the whole problem"? I'm not sure what you mean by that. Can you tell us what you want to prove as mathematical statements?

Any vector which is normal to n3 is then an eigenvector of S with eigenvalue equal to λ1.

 

[vela]

How about considering the vector space as the direct sum of the subspace spanned by $n_3$ and the subspace of vectors perpendicular to $n_3$?

 

[Passers_by]

How about considering the vector space as the direct sum of the subspace spanned by $n_3$ and the subspace of vectors perpendicular to $n_3$?

$n_3$ is a eigenvector with eigenvalue equal to $\lambda_3$.Because $\lambda_1 \ne \lambda_3$, so there is a eigenvector $n_1$ with eigenvalue equal to $\lambda_1$ and $n_1 \perp n_3$.
Let $v_2 \perp n_1$ and $v_2 \perp n_3$,next we can prove $v_2$ is a eigenvalue too. For a symmetric tensor S, $$n_1 \cdot \left( S~ v_2 \right)=\left( S~n_1 \right) \cdot v_2=\lambda_1~n_1 \cdot v_2=0,$$ so $\left( S~v_2 \right) \perp n_1$. In the same way, $\left(S~v_2\right) \perp n_3$. That is $S~v_2=k v_2$，it's also an eigenvector. But what is its eigenvalue, $\lambda_1$ or $\lambda_3$?
That's all I can do.

 

[Gaussian97]

You have been given already 4 ideas/hints on how to proceed, but you have not shown any effort to even consider them, since what you have written is exactly the same as you had in the first post.

 

[Passers_by]

Expand $v_2$ in an orthogonal base formed by eigenvectors: $n_1,n_2,n_3,$$$v_2=k_1n_1+k_2n_2+k_3n_3,$$$$Sv_2=\lambda_1\left(k_1n_1+k_2n_2\right)+\lambda_3k_3n_3,$$but the subspace spanned by $n_1,n_2$ and the subspace by $n_3$ do not intersect, so either $k_1=k_2=0$ or $k_3=0$.
if $k_1=k_2=0,~k_3 \neq 0$, so $v_2$ is spanned by $n_3$. This contradicts that $v_2\perp n_3$. Which means $v_2$ is spanned by $n_1\rm{~and~}n_2$ with eigenvalue equal to $\lambda_1$.

 

[Gaussian97]

That looks promising, but I think that the part:

but the subspace spanned by $n_1,n_2$ and the subspace by $n_3$ do not intersect, so either $k_1=k_2=0$ or $k_3=0$.

is either wrong or you're making some hidden assumptions... Can you elaborate more on this statement?

Alternatively, once you expand $v_2$, can you impose some conditions on the coefficients $k$ before applying $S$?

 

[Passers_by]

That looks promising, but I think that the part:

is either wrong or you're making some hidden assumptions... Can you elaborate more on this statement?

Alternatively, once you expand $v_2$, can you impose some conditions on the coefficients $k$ before applying $S$?

I rewrote the proof as follows:
Expand $n_3$ with eigenvalue equal to $\lambda_3$ in an orthogonal base formed by eigenvectors: $n_1^*,n_2^*,n_3^*$($n_1^*,n_2^*$with eigenvalue equal to $\lambda_1,n_3^*$ with eigenvalue equal to $\lambda_3$)$$n_3=k_1n_1^*+k_2n_2^*+k_3n_3^*,$$ $$Sn_3=\lambda_1\left(k_1n_1^*+k_2n_2^*\right)+\lambda_3k_3n_3^*,$$but the subspace spanned by $n_1^*,n_2^*$ and the subspace by $n_3^*$ do not intersect, so $k_1=k_2=0$, $n_3$ and $n_3^*$ are linearly dependent. Any vector with eigenvalue equal to $\lambda_3$ is linearly dependent. Which means the eigenvalue of eigenvector $v_2$ is $\lambda_1$.
Any vector $v$ which is normal to n3 is a vector of subspace spanned by $n_1$ and $v_2$. Let $v=sn_1+tv_2$,$Sv=S\left( sn_1+tv_2\right )=\lambda_1\left( sn_1+tv_2\right )=\lambda_1 v$.

 

[Gaussian97]

I think all the main ingredients are there, but I find it in general very confusing, skipping some important detail while doing other things that I can't see how they fit in the general prof. Could you organize your ideas to write a complete detailed proof?

 

[vela]

Expand $v_2$ in an orthogonal base formed by eigenvectors: $n_1,n_2,n_3,$$$v_2=k_1n_1+k_2n_2+k_3n_3,$$

Try calculating $n_3 \cdot v_2$.

 

[Passers_by]

I think all the main ingredients are there, but I find it in general very confusing, skipping some important detail while doing other things that I can't see how they fit in the general prof. Could you organize your ideas to write a complete detailed proof?

Here is my idea of proof. First, prove that $v_2$ is an eigenvector. You can see that in #13. And then let's prove that its eigenvalue is $\lambda_1$. And to prove that, I found that any eigenvector with an eigenvalue of $\lambda_3$ is linearly dependent. So $\lambda_3$ is not the eigenvalue of $v_2$. That's what the #17 is about. So $n_1$ and $v_2$ span the subspace of the eigenvectors that have eigenvalue $\lambda_1$. Any vector which is normal to n3 is a vector of this subspace. So we've done the proof.

 

[Passers_by]

Try calculating $n_3 \cdot v_2$.

$n_3\cdot v_2=k_3$. But there's something wrong with that proof in #15. Please don't be bothered by that.

 

[hutchphd]

No. By choice you demand $k_3=0$ which insures $v_2$ orthogonal to $n_3$ . In two lines you can now show desired result. Let S operate on $v_2$

 

[Passers_by]

No. By choice you demand $k_3=0$ which insures $v_2$ orthogonal to $n_3$ . In two lines you can now show desired result. Let S operate on $v_2$

I might know what you mean. If $v_2=k_1n_1+k_2n_2$, then $Sv_2=\lambda_1v_2$. I knew that from the beginning. And it all comes with a premise: there are two eigenvectors($n_1$, $n_2$) that are orthogonal to each other in the subspace in which any vector is normal to $n_3$ and their eigenvalues are equal to $\lambda_1$.

 

[hutchphd]

You have shown then that for any values $k_1,k_2$ you get get an eigenvector with value $\lambda_1$. So choose $k_1=0$ QED what else do you need to show?.

 

[Gaussian97]

Here is my idea of proof. First, prove that $v_2$ is an eigenvector. You can see that in #13. And then let's prove that its eigenvalue is $\lambda_1$. And to prove that, I found that any eigenvector with an eigenvalue of $\lambda_3$ is linearly dependent. So $\lambda_3$ is not the eigenvalue of $v_2$. That's what the #17 is about. So $n_1$ and $v_2$ span the subspace of the eigenvectors that have eigenvalue $\lambda_1$. Any vector which is normal to n3 is a vector of this subspace. So we've done the proof.

Ok, I think that's correct.
I still believe that your mathematical formulation of the proof is very confusing and you should consider rewriting the whole proof in a clearer way. But anyway, I think that at least you got the general idea of the proof.

 

[vela]

$n_3\cdot v_2=k_3$. But there's something wrong with that proof in #15. Please don't be bothered by that.

I think some of us here are a bit confused by what you're looking for since some of the stuff you insist on proving we think is obvious. For instance, if you are given $v_2 \perp n_3$, most here would automatically say $v_2$ is some linear combination of the other two eigenvectors; therefore, the conclusion to be proved follows trivially. As you have said, "you knew that from the beginning," so it wasn't clear what exactly you thought needed to be proved.

You, however, started with $v_2 = k_1 n_1 + k_2 n_2 + k_3 n_3$ (which is fine) and set out to show $k_3 = 0$. To that end, if $v_2 \perp n_3$, you get $n_3 \cdot v_2 = k_3 = 0$.

 

[Passers_by]

I think some of us here are a bit confused by what you're looking for since some of the stuff you insist on proving we think is obvious. For instance, if you are given $v_2 \perp n_3$, most here would automatically say $v_2$ is some linear combination of the other two eigenvectors; therefore, the conclusion to be proved follows trivially. As you have said, "you knew that from the beginning," so it wasn't clear what exactly you thought needed to be proved.

You, however, started with $v_2 = k_1 n_1 + k_2 n_2 + k_3 n_3$ (which is fine) and set out to show $k_3 = 0$. To that end, if $v_2 \perp n_3$, you get $n_3 \cdot v_2 = k_3 = 0$.

The reason for the confusion is that my original idea was to prove that the subspace normal to $n_3$ is a eigenspace with eigenvalue equal to $\lambda_1$. I'm not taking this as a given premise.

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[hutchphd]

Every time one wishes to examine a new idea we are not required to prove Euler's Identity. Symmetric matrices (Hermitian if they are complex) have been extensively examined, and the properties of eigenvectors are well vetted. It is not a useful exercise to reinvent the wheel. This of course requires you to be aware of the existence of the wheel

 

[Gaussian97]

It is not a useful exercise to reinvent the wheel.

I couldn't disagree more with that. A great part of studying mathematics (or physics or any science in general) is to reinvent the wheel, look how professionals construct their wheels until you are able to invent a completely new wheel by yourself.

Why do you think it's better to start by assuming that "Any symmetric matrix diagonalize in an orthonormal base."
Than start by just assuming "Any symmetric matrix diagonalize."?

 

[hutchphd]

Perhaps you misunderstand me. You do need to understand how to build a wheel. But you do not need to recapitulate that knowledge every time a wheeled vehicle is mentioned. Knowing what is common knowledge (among physicists) is also part of the education process. Not having this knowledge means you cannot effectively and efficiently communicate. This was an "advanced physics" post...