Number of eigenvectors for Hermitian matrices

[haisydinh]

Hello,

I am currently trying to study the mathematics of quantum mechanics. Today I cam across the theorem that says that a Hermitian matrix of dimensionality $n$ will always have $n$ independent eigenvectors/eigenvalues. And my goal is to prove this. I haven't taken any linear algebra classes so my knowledge in this field is quite limited.

Now, I have realized that if I were to prove this statement, I would basically have to show that there is a polynomial associated with the matrix and that this polynomial can be factored into a product of distinct linear factors. Mathematically, if I let $M$ to be the Hermitian matrix, $I$ be the identity matrix, and $\lambda$ be the placeholder for the eigenvalues, then I have to show that the following is true for all $n$:

$det(M-I\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2)...(\lambda - \lambda_n)$ (where $\lambda_i$ are independent eigenvalues)

Here, I’m just stuck. I really don’t know how to tackle this at all. If someone could help me out on this problem, it would be really great. Or if you guys can point me to some sort of literature, that would also be good for me. Thank you very much in advance! :)

 

[ShayanJ]

det(M-I\lambda)=0 \Rightarrow (\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)...(\lambda-\lambda_n)=0
This is an algebraic equation of degree n, which according to the fundamental theorem of algebra, has n roots. But these roots need not be distinct. Any number of them may be equal and so there may be degeneracy.

 

[haisydinh]

det(M-I\lambda)=0 \Rightarrow (\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)...(\lambda-\lambda_n)=0
This is an algebraic equation of degree n, which according to the fundamental theorem of algebra, has n roots. But these roots need not be distinct. Any number of them may be equal and so there may be degeneracy.

Hi, thanks for replying.

However, the theorem says that "the polynomial associated with a Hermitian matrix always has $n$ distinct roots". In other words, this is a special property of a Hermitian matrix. I can easily prove this with a $2×2$- Hermitian matrix or a $3×3$-Hermitian matrix, however I want to find a generalized proof to show that this is true for all $n$.

Thanks.

 

[ShayanJ]

I don't know what you did but that's wrong!
A 2\times 2 Hermitian matrix is of the form \left( \begin{array}{lr} a \ \ \ \ \ \ c+id \\ c-id \ \ \ \ \ \ b \end{array} \right). This matrix has two eigenvalues and these two become equal when a-b=\pm 2 i \sqrt{c^2+d^2}. So I'm showing you an infinite number of counter-examples to that statement.

 

[haisydinh]

I don't know what you did but that's wrong!
A 2\times 2 Hermitian matrix is of the form \left( \begin{array}{lr} a \ \ \ \ \ \ c+id \\ c-id \ \ \ \ \ \ b \end{array} \right). This matrix has two eigenvalues and these two become equal when a-b=\pm 2 i \sqrt{c^2+d^2}. So I'm showing you an infinite number of counter-examples to that statement.

Maybe I should be more clear with my question. Let's say that $M = \begin{pmatrix} a & c+id \\ c-id & b \end{pmatrix}$ is our Hermitian matrix, then the polynomial is:

$det(M - I\lambda) = det\begin{pmatrix} a-\lambda & c+id \\ c-id & b-\lambda \end{pmatrix}$

$= \lambda^2 + (a+b)\lambda + (ab-c^2-d^2) = 0$

The discriminant is thus $(a+b)^2 - 4(ab-c^2-d^2) = a^2-2ab+b^2+4(c^2+d^2)=(a-b)^2+4(c^2+d^2)$ which is obviously positive (unless all $a, b, c, d = 0$ which is uninteresting). Therefore the polynomial has 2 distinct roots for $\lambda$. Now, the proof for $3×3$-matrix is more complicated, but will give the same result. So my question is how do we prove this statement for all $n$?

Thank you!

 

[ShayanJ]

The discriminant is thus (a+b)2−4(ab−c2−d2)=a2−2ab+b2+4(c2+d2)=(a−b)2+4(c2+d2)(a+b)^2 - 4(ab-c^2-d^2) = a^2-2ab+b^2+4(c^2+d^2)=(a-b)^2+4(c^2+d^2) which is obviously positive (unless all a,b,c,d=0a, b, c, d = 0 which is uninteresting)

This part is wrong. If you set the discriminant equal to zero, you'll get the equation I mentioned in my previous post! Which means its is possible for the discriminant to be zero without a,b,c and d all be zero!

EDIT: Because a and b are real, that equation can only be true when c and d are zero which means a=b. So the only 2\times 2 matrix with degenerate eigenvalues is the identity matrix. For higher dimensions, there may be more!

 

[haisydinh]

This part is wrong. If you set the discriminant equal to zero, you'll get the equation I mentioned in my previous post! Which means its is possible for the discriminant to be zero without a,b,c and d all be zero!

EDIT: Because a and b are real, that equation can only be true when c and d are zero which means a=b. So the only 2\times 2 matrix with degenerate eigenvalues is the identity matrix. For higher dimensions, there may be more!

Yes, you are absolutely right! How silly of me! Thanks for pointing that out!

Well, it's clear that my proof is wrong here. But the problem still remains though. Maybe it's more relevant to physics rather than to mathematics. Maybe in quantum mechanics, they are talking about a different type of Hermitian matrices that disallow $a$ and $b$ to be the same. In any case, I would have to look more into the context of the physics situation, rather than generalizing it to all Hermitian matrices.

Thanks again!