Linear Algebra minimal Polynomial

Question

Let $T$ be a mapping from $\def\R{\mathbf R}\R^4 \to \R^4$ such that the null space of $T$ is $\{(x,y,z,w):x+y+z+w=0\}$ and the rank of $(T-4I)$ is $3$, and the minimal polynomial of $T$ is $x(x-4)^a$, then $a=\mathord?$

Answer 1

Actually the given two conditions are not needed simultaneously.

From the second condition : Geometric multiplicity of the eigen value $4$ is $$\text{order}(T-4I)-\text{rank}(T-4I)=1.$$ So, there are exactly one Jordan block corresponding to the eigen value $4$. So $(x-4)$ is exactly one factor of the minimal polynomial of $T$ (i.e. , $4$ is a root of the minimal polynomial of multiplicity $1$). So $a=1$.

From the first condition : Given , $N(T)=3$. So , $T$ has three $0$ eigen values. So , characteristic polynomial of $T$ is $x^3(x-4)$. From this and your given minimal polynomial it is clear that $a=1$.