# Diagonalising a matrix with repeated roots.

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?

• If $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $v=\begin{bmatrix}1\\1\end{bmatrix}$ then $v$ is not an eigenvector of $A$. (Calculate $Av$...) Commented Jul 15, 2022 at 10:32
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.