Minimal polynomial of the operator $T:V\oplus W\to V\oplus W$ Let $V$ and $W$ be two finite dimensional vector spaces over $\Bbb R$ and let $T_{1}:V\to V$ and $T_{2}:W\to W$ be two linear transformations whose minimal polynomials are given by $f_{1}(x)=x^{3}+x^{2}+x+1$ and $f_{2}(x)=x^{4}-x^{2}-2$.
Let, $T:V\oplus W\to V\oplus W$ be the linear transformation defined by 
$T(v,w)=(T_{1}(v),T_{2}(w))$ for $(v,w)\in V\oplus W$ and let $f(x)$ be the minimal polynomial of $T$. Then which of the followings are correct?
(1) $\deg(f(x))=7$
(2) $\deg(f(x))=5$ 
(3) $\mathrm{nullity}(T)=1$  
(4) $\mathrm{nullity}(T)=0$ .  
 A: Although the currently accepted answer says otherwise, the polynomials $f_1$ and $f_2$ are not coprime; applying the Euclidean algorithm shows that $\gcd(x^3+x^2+x+1,x^4-x^2-2)=x^2+1$. It then follows that the minimal polynomial$~f$ of$~T$, which is the least common multiple of $f_1$ and $f_2$ as it is the least degree monic polynomial whose evaluation in the restrictions $T_1$ and $T_2$ of$~T$ to $W_1,W_2$ are both zero, equals $x^5+x^4-x^3-x^2+2x+2$. This polynomial having nonzero constant term means that $0$ is not an eigenvalue of$~T$ (root of$~f$), so $T$ is invertible and has nullity$~0$. So answers (2),(4) are correct. 
A: Hint


*

*The minimal polynomial of $T$ is the least common multiple of $f_1$ and $f_2$. Notice that $f_1$ and $f_2$ have not a common root so they are coprime and the minimal polynomial of $T$ is $f_1f_2$.

*$\operatorname{nullity}(T)\ne0$ if and only if $0$ is a root of the minimal polynomial of $T$ if and only if $0$ is a root of $f_1$ or $f_2$ which isn't the case.

