How to create a counter example of a matrix with non orthogonal eigenvector?

Question

I am trying to understand this

orthogonal eigenvectors

$$T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$

let

$$u = \begin{pmatrix} 1 \\ 0 \\ 1 \\ \end{pmatrix}$$

$$v = \begin{pmatrix} 0 \\ 1 \\ 1 \\ \end{pmatrix}$$

They are all linearly independent. $u$ and $v$ are eigen vectors of $T$ right?

I just want to make sure I am correct. So those are samples of eigen vectors that not orthogonal. The catch is the eigen value are the same. Namely 1. How can I create eigen vectors that are not orthogonal and have different eigen value?

I want a counter example of a matrix (doesn't have to be symmetric or diagonal) that have non orthogonal eigen vectors with different eigen value.

Answer 1

For a non-symmetric matrix take:

$$A=\begin{pmatrix} 1& 1 \\ 2& 2 \end{pmatrix}. $$

Then $(1,2)^T$ is a eigenvector to eigenvalue $3$ and $(1,-1)^T$ is a eigenvector to eigenvalue $0$ but they are not orthogonal.