# Prove if $\mu(x)$,$\mu_1(x)$,$\mu_2(x)$ are minimal polynomial of $T$,$T|_{V_1}$,$T|_{V_2}$ then…

Let $T:V\rightarrow V$ a linear operato above a vector space of finite dimension and a descomposition in direct sum $V=V_1\oplus V_2$, with $V_1$ and $V_2$ subspaces $T-stable$ of $V$. Prove if $\mu(x)$,$\mu_1(x)$,$\mu_2(x)$ are minimal polynomial of $T$,$T|_{V_1}$,$T|_{V_2}$ then, $\mu(x)$ is the Least common multiple in $\mu_1(x)$ and $\mu_2(x)$

My work:

As $V=V_1\oplus V_2$ and $T:V\rightarrow V$ then we can decomposite $T$ in $T=T_1 \oplus T_2$

Let $B_1=\{u_1,...,u_n\}$ and $B_2={w_1,...,w_m}$ basis for $V_1$ and $V_2$

Let $A_1$ the matrix associtated to $T_1$ and $A_2$ the matrix associated to $T_2$

As $V=V_1\oplus V_2$ then $B=\{v_1,...,v_n,w_1,...,w_m\}$ is a basis for $V$

Moreover, as $T=T_1 \oplus T_2$ then $T(u_i)=T_1(u_i)$ with $1\leq i\leq n$ this implies $w_j=0$ for $j=1,...,m$ Analogous for $T(w_i)$.

Then, the matrix associated to $T$ is $K=\begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}$

This implies $P_K(x)=det(xI-K)=det\begin{bmatrix} xI-A_1 & 0 \\ 0 & xI-A_2 \end{bmatrix}=det(xI-A_1)det(xI-A_2)=P_{A_1}(x)P_{A_2}(x)$

As the minimal polynomial divide to the characteristics polynomial, then:

$\mu_1(x)q_1(x)\mu_2(x)q_2(x)=\mu(x)$ with $q_1(x)\,,q_2(x)\in\mathbb{K}{[x]}$

In consequence, $\mu_1(x)\mu_2(x)S(x)=\mu_(x)$ where $S(x)=q_1(x)q_2(x)$

I dont'sure if my proof is correct, and my supposition of $T=T_1 \oplus T_2$

Can someone help me? Thanks.

The minimal polynomial $\mu(t)$ can be equivalently determined as the generator of the ideal of $k[t]$ consisting of those polynomials $p(t)$ with $p(T)=0$. Now given the decomposition as above, it is clear that $p(T)=0$ if and only if both $p(T_1)=0$ and $p(T_2)=0$, so that the ideal generated by $\mu(t)$ is the intersection of the ideals generated by $\mu_1(t)$ and by $\mu_2(t)$. And the generator of an intersection of ideals is the lcm of their generators, practically by definition of lcm.