Let $T:\mathbb R^4\to \mathbb R^4$ be a linear map such that null space of T is

$$\{(x,y,w,z)\in \mathbb R^4 :x+y+w+z=0\}$$

and the rank of $(T-4I_4)$ is 3 .

If the minimal polynomial of $T$ is $x(x-4)^{\alpha}$, then $\alpha=?$

My thought:

Here nullity of T is 3$\implies $rank(T)=1 .So one non zero column vector.

rank $(T-4I_4)$ = 3 $\implies$ nullity $ (T-4I_4)=1 \implies4$ is an eigen value of T .

Therefore I found 0 and 4 as eigen values .Does there exist other eigen values ?If not ,then I know the characteristic polynomial is of the form $x^m(x-4)^n$ and minimal polynomial divides the characteristic polynomial. I can't further proceed. Please help.

Thanks for your time.


Since the nullity of $T$ is $3$, therefore $0$ is an eigen value of $T$, moreover the eigenspace corresponding to $\lambda=0$ is of dimension $3$. Thus the multiplicity of $\lambda=0$ is at least $3$. But given the fact that $4$ is also an eigen value of $T$ you can conclude that the characteristic polynomial can only be of the form $\lambda^3(\lambda-4)$ (and no other eigenvalues). Now minimal has to divide characteristic. So $\alpha=1$.


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