TITLE: Eigenspace. What is it?
QUESTION [29 upvotes]: What is an eigenspace?
No video or anything out there really explains what an eigenspace is?
From what i have understood it is just a direction. But why do we need it?.
The following questions have been bugging me for quite a while, and I can't find a real straight forward answer to them. Hopefully one of you can help me.

What is an eigenspace?
Why are the eigenvectors calculated in a diagonal?
What is the practical use of the eigenspace? Like what does it do or what is it used for? other than calculating the diagonal of a matrix.
why is it important o calculate the diagonal of a matrix?

I want to know mainly because I just passed a linear algebra course and I have no idea what eigenspace is. Which is embarrassing for me and for my professor because he didn't explain what they were, he just basically said: "This is how you calculate it and if you want to know more then read about it in the book ".

REPLY [3 votes]: To make a long story short: Given a linear map $L:\>V\to V$ and any complex number $\lambda$ one considers the set $E_\lambda\subset V$ of all vectors $x\in V$ that are just stretched by the factor $\lambda$ under $L$:
$$E_\lambda:=\{x\in V\>|\> Lx=\lambda x\}\ .$$
This set is obviously a subspace, but (in a finite-dimensional setting) only for a handful of $\lambda$-values it is interesting, that is to say: of positive dimension. These special $\lambda$s are called eigenvalues of the given $L$ and together form the set ${\rm spec} (L)\subset{\mathbb C}$.
It turns out that (a) the nontrivial $E_\lambda$  are all-important in the understanding of the map $L:\>V\to V$ and that (b) working with vectors $e_\lambda\in E_\lambda$, $\>\lambda\in{\rm spec} (L)$, as basis vectors greatly simplifies computations involving $L$.
An example: When $L$ is the differentiation operator ${d\over dt}$ on functions $f:\>{\mathbb R}\to{\mathbb C}$ it pays to  use the functions $e_\lambda:\ t\mapsto e^{\lambda t}$ as "basis functions", because ${d\over dt}e_\lambda =\lambda\>e_\lambda$.