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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.

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user4951
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  • here's that tutorial – Ben Grossmann Jun 06 '16 at 14:30
  • As long as you take diagonalizable symmetrical matrices, the eigenvectors will be orthogonal. – Jean Marie Jun 06 '16 at 14:32
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    Long story short: something went "wrong" with your second example because you repeated an eigenvalue. If you want non-orthogonal eigenvectors, you will need a non-symmetric (diagonalizable) matrix. – Ben Grossmann Jun 06 '16 at 14:33
  • Ah what do you mean by diagonlizable matrix? I thought all matrix are diagonalizeable. Just use normal algorithm we use to solve linear equation. Hmmm? – user4951 Jun 06 '16 at 14:37
  • @Maik Pickl, you are right, but agree with me that identity matrix is so special... – Jean Marie Jun 06 '16 at 14:40
  • @JeanMarie I deleted my comment since the question has changed. But this always happens when you have repeated eigenvalues. You can choose a basis of the eigenspace which does not have to be orthogonal. – Maik Pickl Jun 06 '16 at 14:42
  • I think we can diagonalize any matrix using reduced row echilon form or something like that – user4951 Jun 06 '16 at 14:44
  • @Maik Pickl You are right. Thank you. – Jean Marie Jun 06 '16 at 14:53
  • @JimThio No. The matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ is not diagonalizable. It has the eigenvalue $1$ with algebraic multiplicity $2$ but geometric multiplicity only $1$. – Ian Jun 06 '16 at 14:57

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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.

Maik Pickl
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