# Eigenvectors and inner product

• I
Hi, what is the physical meaning, or also the geometrical meaning of the inner product of two eigenvectors of a matrix?

I learned from the previous topics that a vectors space is NOT Hilbert space, however an inner product forms a Hilbert space if it is complete.

Can two eigenvectors which are linearly independent form a Hilbert space in their inner product?

e_1 = [x_1, y_1]
e_2 = [x_2, y_2]

##\langle e_1, e_2 \rangle## ?

Thanks

Mentor
2022 Award
Hi, what is the physical meaning, or also the geometrical meaning of the inner product of two eigenvectors of a matrix?
An inner product of two vectors, let them be eigenvectors of some transformation or not, is an assignment which can be used to measure lengths and angles, physically and geometric. An explanation and a figure can be found here: https://en.wikipedia.org/wiki/Inner_product_space or more elaborated here: https://arxiv.org/pdf/1205.5935.pdf
I learned from the previous topics that a vectors space is NOT Hilbert space, ...
A Hilbert space is a vector space, but not every vector space is also a Hilbert space. That is: a Hilbert space has more properties.
... however an inner product forms a Hilbert space if it is complete.
Basically, yes, but the wording is a bit weird. Not the inner product forms anything, a Hilbert space is defined by being a real or complex vectorspace equipped with an inner product (additional property). The inner product defines measurements as angles and lengths, e.g. a distance: ##\operatorname{dist}(x,y)=||x-y||=\sqrt{\langle x-y,x-y \rangle}##, which is basically the theorem of Pythagoras here. This distance has to be complete, that is every Cauchy sequence (the elements of the sequence get closer and closer) converges, which means the infinite point towards the elements are narrowing down is actually part of the space. Completeness thus depends on how we measure distances. There is generally more than one way to define a distance. So Hilbert spaces are strictly speaking triplets: (vector space, field (real or complex), inner product).

E.g. with the rational numbers we'll get a problem. Let's define ##a_1=b_1=1\; , \;a_{n+1}=a_n+b_n\; , \;b_{n+1}=a_{n+1}+a_n\; , \;c_n = \dfrac{b_n}{a_n}##. Then ##(c_n)_{n\in \mathbb{N}}## defines a Cauchy sequence of rational numbers which converges to ##\sqrt{2}##. But this limit isn't part of the space ##\mathbb{Q}##, so ##\mathbb{Q}## is incomplete. If we take the exact same situation to the reals (or complex numbers), then this limit is part of the space - ##\mathbb{R}## and ##\mathbb{C}## are complete spaces, i.e. also Hilbert spaces.
Can two eigenvectors which are linearly independent form a Hilbert space in their inner product?

e_1 = [x_1, y_1]
e_2 = [x_2, y_2]

##\langle e_1, e_2 \rangle## ?
This has nothing to do with eigenvectors. Besides, any two vectors are always eigenvectors of some linear function. Two linear independent vectors over the real or complex numbers span a two-dimensional real or complex vector space. E.g. with the Euclidean metric we get a Euclidean space. And this is a two-dimensional Hilbert space, as the real or complex numbers guarantee that any sequence of vectors which is narrowing down by Euclidean distance (Pythagoras) also contains the vector towards the sequence is converging.

• SeM
An inner product of two vectors, let them be eigenvectors of some transformation or not, is an assignment which can be used to measure lengths and angles, physically and geometric. An explanation and a figure can be found here: https://en.wikipedia.org/wiki/Inner_product_space or more elaborated here: https://arxiv.org/pdf/1205.5935.pdf
A Hilbert space is a vector space, but not every vector space is also a Hilbert space. That is: a Hilbert space has more properties.

Basically, yes, but the wording is a bit weird. Not the inner product forms anything, a Hilbert space is defined by being a real or complex vectorspace equipped with an inner product (additional property). The inner product defines measurements as angles and lengths, e.g. a distance: ##\operatorname{dist}(x,y)=||x-y||=\sqrt{\langle x-y,x-y \rangle}##, which is basically the theorem of Pythagoras here. This distance has to be complete, that is every Cauchy sequence (the elements of the sequence get closer and closer) converges, which means the infinite point towards the elements are narrowing down is actually part of the space. Completeness thus depends on how we measure distances. There is generally more than one way to define a distance. So Hilbert spaces are strictly speaking triplets: (vector space, field (real or complex), inner product).

E.g. with the rational numbers we'll get a problem. Let's define ##a_1=b_1=1\; , \;a_{n+1}=a_n+b_n\; , \;b_{n+1}=a_{n+1}+a_n\; , \;c_n = \dfrac{b_n}{a_n}##. Then ##(c_n)_{n\in \mathbb{N}}## defines a Cauchy sequence of rational numbers which converges to ##\sqrt{2}##. But this limit isn't part of the space ##\mathbb{Q}##, so ##\mathbb{Q}## is incomplete. If we take the exact same situation to the reals (or complex numbers), then this limit is part of the space - ##\mathbb{R}## and ##\mathbb{C}## are complete spaces, i.e. also Hilbert spaces.

This has nothing to do with eigenvectors. Besides, any two vectors are always eigenvectors of some linear function. Two linear independent vectors over the real or complex numbers span a two-dimensional real or complex vector space. E.g. with the Euclidean metric we get a Euclidean space. And this is a two-dimensional Hilbert space, as the real or complex numbers guarantee that any sequence of vectors which is narrowing down by Euclidean distance (Pythagoras) also contains the vector towards the sequence is converging.

Thanks for an excellent elucidation.

I basically want to state that two eigenvectors of a matrix form a Hilbert space, and use the following criteria to prove that:

The eigenvectors satisfy the condition

\begin{equation}
\langle x,y \rangle = \int_a^bx(t)\overline{y(t)}dt,
\end{equation}

which is complete, and form the complex vector space ##L^2[a,b]## which satisfies the minimal condition:

\begin{equation}
x(t)\overline{x(t)}=|x(t)|^2.
\end{equation}

However, I want to make sure I am not writing something odd here, or even connecting two things that are not connected. As from your reply, I understand that one can conclude that two vectors form a Hilbert space, but whether it is complete or not, requires the continuous nature of these, so to satisfy what you describe here as the Cauchy criteria. Is it correctly given here, with the "minimal condition"?

Thanks!