I have a linear application $\phi : \mathbb R^6 \rightarrow \mathbb R^6$, and I have given its minimal polynomial : $t^2 (t^2+1)(t-1)$.
how can I find the dimension of the image of phi?
Sorry if the question is trivial, i'm still learning :) I thought that maybe there is a link with the dimension of the eigenspaces but I figure this out...
From the polynomial, all its eigenvalue are $i,-i,1,0$. Consider its canonical form, there are two possible answers 4 or 5. the former one means an extra eigenvector to $0$, the latter to $1$. The following are two possible matrix.The first 2-dimensional block is for $x^2+1$, $1$ in the third row for $x-1$, the next 2-dimensional block for $x^2$. the last row can be eigenvalue $0$ or $1$ and can not be $i$ or $-i$ because in real form, imaginary eigenvalue will appear with a pair.
\begin{equation*} \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right ] \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right ] \end{equation*}