Let be $T:\mathbb{R}^7\rightarrow \mathbb{R}^7$
Such that $(T-15I)^3=0$ and $\dim\text{Im}(T-15I)^2=2$ find the Jordan normal form of $T$
If $(T-15I)^3=0$ so the minimal polynomial can be $(T-15I)^3=0$,$(T-15I)^2$,$(T-15I)$ But because $\dim\text{Im}(T-15I)^2=2$ So it can not be $(T-15I)^2$ and $(T-15I)$ it had to be $\dim\text{Im}(T-15I)^2=7$
The Characteristic polynomial is $(T-15I)^7$.
So we have $B_1$ of size $3$ and we are left with 7-3=4 eigenvalues, how can I find the reminding jordan normal form blocks?
The largest Jordan block size is $3$ because $(T-15I)^{3}=0$, and there have to be two of those because the rank of $(T-15I)^{2}$ is $2$. Because the dimension of the space is $7$, then there is also one Jordan block of size $1$.