Finding eigenvalues and eigenvectors geometrically. Let $A=\left[\begin{matrix} 0 & 1\\1 & 0 \end{matrix}\right]$ represent reflection about the line $y=x$.
I can calculate eigenvalues and eigenvectors mathematically, but I have hard time getting the results geometrically. There are some similar posts available but they are not of any help. For example in this post, I don't know how they found 1 and -1 as eigenvalues?  I must be missing some points. So the question is,
how to find eigenvalues and eigenvectors geometrically? Why in this case eigenvalues are 1 and -1? and how to obtain corresponding eigenvectors?
 A: $A=\left[\begin{matrix} 0 & 1\\1 & 0 \end{matrix}\right]$
Consider the linear map associated with $A$ ,
$T_A:\Bbb{R}^2\to\Bbb{R}^2$ defined by $T_A(x, y) =(y, x) $
$\lambda$ is an eigenvalue of $T$ implies $\exists v\in\Bbb{R^2}\setminus\{0\}$ such that $\text{span}\{v\}$ is $1$- dimensional invariant subspace of $\Bbb{R^2}$ i.e for any vector $u\in \text{span}\{v\} , Tu\in\text{span}\{v\}$
A one dimensional subspace of $\Bbb{R}^2$ is a line through origin.
Hence you have to find  a line through origin such that for any point $(x,y)\in\Bbb{R}^2$ on that line, after applying the transformation $T$ i.e after switching the co-ordinate  the point still lie on that line.
Choose the line $y=x$ i.e the subspace $\{(x,x):x\in\Bbb{R}\}$ . Now it's easy to see after switching the co-ordinate it will be still one the line $ y=x$ . In that case eigenvalue is $1$.
There is another one $, y=-x$ i.e the subspace $\{(x,-x):x\in\Bbb{R}\}$ . Then after switching the co-ordinate any points will be in the same line but change it's direction ( correspond negative eigenvalue $-1$)
