I am told that $f(A)=g(A)h(A), f(A)=0$ is a polynomial of an $n \times n$ matrix $A$.
I am further told that $g(x)$ and $h(x)$ are coprime. I need to show that $\mathbb{R}^n$ is the direct sum of $\text{ker }g(A)$ and $\text{ker }h(A)$.
My attempt at a solution: Since $g(x)$ and $h(x)$ are coprime, this means that they have no roots or factors in common, so I can deduce that $\text{ker }g(A) \cap \text{ker }h(A) ={0}$.
I now want to show that if a vector $v \in \mathbb{R}^n$ is not in $\text{ker }g(A)$, then it is in $\text{ker }h(A)$ and vice versa.
Since $f(A)=0$, we have $f(A)v=0 \Rightarrow g(A)h(A)v=0$
Because of commutativity of polynomial multiplication, this also means that $h(A)g(A)v=0$.
If $v \notin \text{ker }g(A)$ then $g(A)v\ne 0$ so $h(A)v=0$. Is this true? Am I on the right track?
Any help would be greatly appreciated!
.Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v \in \Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that : $$ h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \\ g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0 $$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.