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?

• Draw a few vectors and their images. Try to understand when it's the case that a vector and its image have the same direction (we count the exact opposite direction as the same direction too; it corresponds to a negative eigenvalue). Those vectors are eigenvectors. Commented Jul 15, 2022 at 15:41
• A reflection can't change lengths, so its eigenvalues can only be $1$ and $-1$. The eigenvector corresponding to $1$ is a vector left unchanged by the reflection, while the one corresponding to $-1$ gets turned completely around by the reflection. Commented Jul 15, 2022 at 16:00

$$A=\left[\begin{matrix} 0 & 1\\1 & 0 \end{matrix}\right]$$

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 $$-1$$)

• why did you not pick the subspace {(-x, x : x in R)}? In this case direction vector would be different, i.e. v= (x, -x) Commented Jul 17, 2022 at 20:29
• $U_1=\{(x,-x):x\in \Bbb{R}\}$ and $U_2=\{(-x,x):x\in \Bbb{R}\}$ represent the same subspace. They are the same line. Draw it. A subspace is closed under scalar multiplication. $(x, -x) \in U_1$ implies $-1\cdot (x, -x) =(-x, x) \in U_1$ Commented Jul 19, 2022 at 8:23