I have been looking at diagonalising some 2x2 matrices recently. Often when there are repeated roots it is clear that I can't form independent eigenvectors. I then wanted to consider two specific matrices with repeated roots to help me think about when it is possible/impossible to diagonalise. When I did this I was a bit unclear as to why one situation led to linearly independent eigenvectors and the other didn't.
The identity matrix \begin{bmatrix}1&0\\0&1\end{bmatrix} must be diagonisable (as it is already in that form!). The characteristic equation has repeated root 1, leaving us with \begin{bmatrix}0&0\\0&0\end{bmatrix} which leads to $0x+0y=0.$ I think this means that I can use any two independent eigenvectors to begin the diagonalization process, i.e. I am free to choose almost any values for x and y when defining the eigenvectors.
I then tried this again with \begin{bmatrix}1&1\\0&1\end{bmatrix} which led to the equations
1)$0x+y=0$
2)$0x+0y=0$
Why can I not form two linearly independent eigenvectors from here? (I understand that this matrix is not diagonalisable...) For example, why do \begin{bmatrix}1\\0\end{bmatrix} and \begin{bmatrix}1\\1\end{bmatrix} not work?
Thanks in advance.
Yes, any two linearly independent vectors will work for the identity matrix.
On the other hand, in the case of $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$, you got the system$$\left\{\begin{array}{l}y=0\\0=0,\end{array}\right.$$whose solutions are the vectors of the form $(x,0)$ ($x\in\Bbb R$). And any two vectors of that for are linearly dependent.