Null space of $T$ is {$(x,y,z,w)\in R^4:x+y+z+w=0$}. Rank of $(T-4I_4)$ is 3. Then minimal polynomial of $T$ is $x(x-4)$. Let $T:R^4\to R^4$ be a linear map. Null space of $T$ is {$(x,y,z,w)\in R^4:x+y+z+w=0$}. Rank of $(T-4I_4)$ is 3.The minimal polynomial of $T$ is $x(x-4)^a$. Then prove that $a=1$.
 A: Since the minimal polynomial vof $T$ is $x(x-4)^a$, we have that $T$ has exactly the eigenvalues $0$ and $4$ and the char. polynomial of $T$ is $x(x-4)^3$. Then we have
$\mathbb R^4=ker(T) \oplus ker((T-4I_4)^3)$
hence 
$4= \dim ker(T)+ \dim ker((T-4I_4)^3)=3+\dim ker((T-4I_4)^3)$,
thus
$\dim ker((T-4I_4)^3)=1$, which shows that $ker((T-4I_4)^3)=ker(T-4I_4).$
Consequently: $\mathbb R^4=ker(T) \oplus ker(T-4I_4)$.
This gives $a=1$
A: Nullity of $A$ =$3$ as you can see from the definition.
Nullity of A is precisely the Geometric multiplicity of $0$. So that gives you G.M of $0= 3$  
Given is rank of $(T-4I)=3$. This implies Nullity of $(T-4I)=1$ ,which is the Geometric multiplicity of $4$ is $1$.
So in total you get $4$ linearly independent eigenvectors i.e 3 eigenvectors corresponding to $0$  and $1$ eigenvector corresponding to $4$.
And if a linear operator has the number of linearly independent eigenvectors equal to the dimension of the domain(in this case $4=dim(R^4)$) then the operator is diagonalizable.hence it's minimal polynomial is product of linear or irreducible factors.Hence $a=1$
