Can a 2d matrix have a 1d vector space?

Solution 1:

Note that the matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ has characteristic equation given by $\lambda^{2}=0$ so the eigenvalue is given by $\lambda=0$ with algebraic multiplicity $2$. By other hand the eigenspace is given by ${\rm span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$. Hence the geometric multiplicity is $1$. We have that algebraic multiplicy is not equal to geometric multiplicity, hence the matrix $ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ is not diagonalizable in that sense. But we can considerate the generalised eigenvectors and then we can find the Jordan descomposition $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^{-1}$.Hence in the Jordan sense the square matrix is diagonalizable. I am not a physicist, but in mathematics this means that in Jordan's sense it is possible to sum the generalized eigenspaces to create the corresponding Jordan diagonal matrix. That is "the matrix is not defective".