How Do Eigenvalues and Eigenvectors Connect to Fourier Transforms?

[MrAlbot]

Hello guys, is there any way someone can explain to me in resume what eigen values and eigenvectors are because I don't really recall this theme from linear algebra, and I'm not getting intuition on where does Fourier transform comes from.

my teacher wrote:
A$\overline{v}$ = λ$\overline{v}$

then he said that for a vector $\overline{x}$

$\overline{x}$ = $\sum^{n}$ $_{i=1}$ x_i $\overline{e}$_i

and he calls this $\overline{e_i}$ the inicial ortonormal base

the he says that this is equal to

$\overline{x}$ = $\sum^{n}$ $_{i=1}$ $\widehat{x}$_i $\overline{v}$_i

where $\overline{v}$_i is the base of the eigenvectors of A


then he says that y=A$\overline{x}$

$\overline{y}$ = $\sum^{n}$ $_{i=1}$ y_i $\overline{e}$_i = $\sum^{n}$ $_{i=1}$ $\widehat{y}$_i $\overline{v}$_i = A$\sum^{n}$ $_{i=1}$ $\widehat{x}$_i $\overline{v}$_i = $\sum^{n}$ $_{i=1}$ A$\widehat{x}$_i $\overline{v}$_i = itex]\sum^{n}[/itex] $_{i=1}$ $\widehat{x}$_i A $\overline{v}$_i = as A$\overline{v}$_i is λ$\overline{v}$_i = itex]\sum^{n}[/itex] $_{i=1}$ $\widehat{x}$_i λ_i $\overline{v}$_i

So we get that $\widehat{x}$_i λ = $\widehat{y}$_i

I Would like to know the intuition behind this and how it relates to the Fourier Series/ Fourier Transform.
I'd really apreciate not to go into deep mathematics once I have very very weak Linear Algebra bases and I will have to waste some time relearning it, but unfortunately I don't have time now.


Hope someone can help!

Thanks in advance!

Pedro

 

[UltrafastPED]

In linear algebra you will have "diagonalized the matrix" towards the end of the term; this process finds the eigenvalues (the terms on the diagonal) and the eigenvectors (the new set of basis vectors for the system).

Thus if you can diagonalize the matrix, a complete set of eigenvectors will exist; they have very nice analytical properties. They correspond to the physical "modes of the system" - if you bang something in the same direction as one of its eigenvectors, then it will only respond in that direction; if you hit it elsewhere, you get multiple responses. That is the significance of the eigenvector equation ... used heavily in acoustics and quantum mechanics, among others.

 

[hilbert2]

Not all matrices have a set of eigenvectors that spans the whole vector space. As an example consider the rotation matrix in ℝ²:

\begin{pmatrix}
cos\theta & -sin\theta \\
sin\theta & cos\theta
\end{pmatrix}

Unless $\theta$ is a multiple of $\pi$, this matrix doesn't have eigenvectors at all!

Usually in applications in physics and engineering, the matrices are hermitian, which guarantees a complete set of eigenvectors.

 

[WWGD]

Mr. Albot:
Maybe you can look at hilbert2's example to illustrate the concept: if 1 were an eigenvalue, then a vector would be sent to itself ( or, more precisely, to the same place (x,y) where it starts; after a rotation). Clearly, like hilbert2 says, 1 can only be an eigenvalue if you rotate by an integer multiple of $$\pi$$ , and the eigenvectors would be all points that are fixed by the rotation. Notice that if $$\lambda=1$$ is an eigenvalue, that means $$Tv=v$$ , so that v is fixed by the transformation.

 

[MrAlbot]

Thanks a lot guys! I just started to study Linear Algebra from the beginning because I wasn't understanding anything you were saying, but only now I can to see how usefull your comments were! Algebra is beautifull ...Thanks a lot again!

 

[WWGD]

A correction to my post #4: that should be an integer multiple of $2\pi$ , not an integer multiple of $\pi$.

 

[MrAlbot]

exactly! that makes a lot more sense now, but I got the point the first time. Do you know where can I find the best place to learn the derivation of Fourier transform? Right now I am learning from khan academy once I'm a little short on time but its being a pleasant trip over linear algebra. How exactly do I map from the R^n to the complex map ?
Best regards

edit: what I really want to know is the derivation of laplace transform and Z transform, once Fourier comes from that.