Show that $\ker(f-a)(f-b)=\ker (f-a)\oplus \ker (f-b)$ where $f$ is a linear map. If $f:V\to V$, is a linear map, I use the notation $(f-a)(x):=f(x)-ax$ and $(f-a)(f-b)$ for $p(f)$ where $p(x)=(x-a)(x-b)\in \mathbb R[x]$, moreover  "multiplication" of linear maps means "composition" (in finite dimension, one might think of $f$ as a square matrix and then it is matrix-multiplication).
Let $V$ a vector space (possibly of infinite dimension). Let $f:V\to V$ a linear map and let $a\neq b$. Prove that $$\ker [(f-a)(f-b)]=\ker(f-a)\oplus \ker(f-b).$$
The fact that $\ker(f-a)+\ker(f-b)$ is a direct sum is clear. Also, for the inclusion $\ker(f-a)\oplus\ker(f-b)\subset \ker[(f-a)(f-b)]$ I did : let $u=v+w$ where $v\in \ker(f-a)$ and $w\in \ker(f-b)$. Then it's easy to prove that $u\in \ker[(f-a)(f-b)]$. 
For the converse inclusion, I have some difficulties.
Let $u\in V$ s.t. $$(f-a)(f-b)(u)=0.$$
Try
Let $u=(u-v)+v$ where $v\in \ker(f-b)$. Then $$0=(f-a)(f-b)(u)=(f-a)(f-b)(u-v)=(f-b)(f-a)(u-v),$$
now, I suppose that $u-v\in \ker(f-a)$, but I'm not sure that the fact that $(u-v)\notin \ker(f-b)$ and $(f-b)(f-a)(u-v)=0$ implies that $u-v\in \ker(f-a).$ 
 A: Remember that, when two subspaces directly sum to the full space, then the decomposition is unique. So, assuming $v$ is just some vector in $\ker(f - b)$, then you might struggle. You need to know exactly what vector it is ahead of time. That's why I often find it's best to assume the conclusion of the result (even though such an assumption is fatuous) and work out exactly what the decomposition must be.
So, suppose $v \in \ker (f - b)(f - a) = \ker (f - a)(f - b)$, and it takes the form $v = u + w$ where $u \in \ker (f - a)$ and $w \in \ker (f - b)$. Then,
\begin{align*}
(f - b)v &= (f - b)u + (f - b)w = (f - b)u \\
&= (f - a)u + (a - b)u \\
&= (a - b)u,
\end{align*}
hence
$$u = \frac{1}{a - b}(f - b)v.$$
This is exactly what I wanted: a formula for the $u$ component, using only $f$ and the vector $v$. Similarly, we can deduce the only possible value for $w$:
$$w = \frac{1}{b - a}(f - a)v.$$
Now, let's reject our fatuous assumptions, and prove that our guess for the decomposition of $v$ is indeed correct. Note that
$$(f - a)u = \frac{1}{a - b}(f - a)(f - b)v = \frac{1}{a - b} 0 = 0,$$
hence $u \in \ker (f - a)$ as required. A similar proof works for $w \in \ker (f - b)$. Now, we just need to show that $u + w = v$. We have,
\begin{align*}
u + w &= \frac{1}{a - b}(f - b)v + \frac{1}{b - a}(f - a)v \\
&= \frac{1}{a - b}((f - b)v - (f - a)v) \\
&= \frac{1}{a - b}(f(v) - bv - f(v) + av) \\
&= \frac{1}{a - b}(a - b) v = v,
\end{align*}
which completes the proof.
A: Hint: for each $x \in V$, $x=(f-b)(\frac{1}{a-b}x)-(f-a)(\frac{1}{a-b}x)$. 
