Direct sum and characteristic polynomial I'm reading the proof of the Jordan Decomposition Theorem, and I'm having trouble understanding the following:
Let $A\in End(V)$ and $\dim V<\infty$. Suppose we have write $V$ as a direct sum of subspaces $M_i$ (with $M_i$ being invariant)
$$V=M_1\oplus\ldots\oplus M_s$$
Why can we deduce from this, that the characteristic polynomial of $A$, is the product of each of the characteristic polynomials of the restrictions $A_{|M_i}$?  Ie. that
$$\chi_{(A)}(x)=\chi_{(A_{|M_1})}(x)\ldots\chi_{(A_{|M_s})}(x)$$
 A: First of all, we're not simply decomposing $V$ into any direct sum of subspaces - that would be far too easy! We're decomposing $V$ into a direct sum of subspaces $M_1 \oplus \dots \oplus M_s$ that are specially chosen so that the image of each $M_i$ under $A$ is contained inside $M_i$. In other words, each $M_i$ is an invariant subspace under the action of $A$.
Now, having established that the $M_i$'s are invariant subspaces under $A$, it makes sense to consider the restrictions $A|_{M_i}$, which are linear maps $M_i \to M_i$, for each $i$. The matrix for the full map $A : V \to V$ can be written in the block diagonal form
$$ A =  \begin{bmatrix} A|_{M_1} & 0 & \dots & 0 \\ 0  & A|_{M_2} & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & A|_{M_s} \end{bmatrix},$$
where each of the blocks are associated with one of the subspaces $M_i$ of $V$.
The characteristic polynomial is then
$$ \begin{eqnarray} \chi_{A}(x)  &=& {\rm det}(A - xI) \\ &=& {\rm det} \begin{bmatrix} A|_{M_1} - x1_{M_1} & 0 & \dots & 0 \\ 0  & A|_{M_2} - x1_{M_2}& \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & A|_{M_s}-  x1_{M_s} \end{bmatrix} \\ &=&  {\rm det}(A|_{M_1} - x1_{M_1}) \times  {\rm det} ( A|_{M_2} - x1_{M_2}) \times \dots \times {\rm det} ( A|_{M_s}-  x1_{M_s} ) \\ &=& \chi_{A|_{M_1}}(x) \times \chi_{A|_{M_2}} (x) \times \dots \times \chi_{A|_{M_s}}(x) \end{eqnarray}  $$
