Prove that $V = \ker T \oplus \text{Im}T$ Let $T:V\to V$ such that $f_T = \sum_{i=0}^n c_ix^i$ and $c_1 = c_n = 1, c_0 = 0$.
Prove that $V = \ker T \oplus \text{Im}T$.
My thoughts so far: 


*

*For some basis $B$, we have $[T]_B = A$. We know that $0=c_0 = (-1)^n\det A$. Therefore, $\det A = 0$ which implies $A$ isn't invertible and it's column are linearly independent.

*$0$ is an eigenvalue of $T$.

 A: I will directly prove that $V$ is the sum of $\ker T$ and $\text{im}T$. Since the characteristic polynomial is given, I will use Caley-Hamilton to proceed while using $c_1=1$, $c_0=0$.
Let $x$ in $V$. Then $f_T(T)x=0$ by Caley-Hamilton, hence
$$
\sum_{i=1}^n c_i T^ix = 0,
$$
or equivalently
$$
T\left(\sum_{i=1}^n c_i T^{i-1} x\right)=T\left(x+\sum_{i=2}^n c_i T^{i-1} x\right)=0,
$$
which helps us to identify an element in the null space of $T$ associated to the given $x$.
Now writing
$$
x = \underbrace{x+\sum_{i=2}^n c_i T^{i-1}x}_{\in \ker T} - \underbrace{\sum_{i=2}^n c_i T^{i-1}x}_{\in \text{im} T}
$$
shows that $\ker T + \text{im} T = V$. 
By the rank–nullity theorem, 
$$
\dim \ker T + \dim\text{im} T  = \dim V = \dim(\ker T + \text{im} T) = \dim \ker T + \dim\text{im} T + \dim(\ker T \cap \text{im} T),
$$
which implies that $\ker T \cap \text{im} T=\{0\}$, and the sum is direct.

The proof that the sum is direct can also be done with Caley-Hamilton. Take $y\in \ker T \cap \text{im} T$. Then there is $x\in V$ such that $Tx=y$ and $T^2x=0$. The latter implies $T^ix=0$ for all $i\ge 2$. Then by Cayley-Hamilton
$$
0=\sum_{i=1}^n c_i T^ix = T x .
$$
Hence $Tx=0$, $y=0$, and  $\ker T \cap \text{im} T=\{0\}$.
A: Since $c_0=0$
$$f_T = \sum\limits_{i=0}^n c_ix^i=x\sum\limits_{i=0}^{n-1} c_{i+1}x^i=xg(x)$$
Since $c_1=1$, $x$ and $g(x)$ are coprime. The minimum polynomial $m_T$ must contain all factors that are divided by $f_T$ and so, $m_T=xh(x)$, where $h|g,\gcd(x,h)=1$. So by theorem of invariant factors for minimal polynomial, $V$ is the direct sum of invariant subspaces of $T$ and $h(T)$, i.e.
$$V=V_1\oplus V_2$$
where $V_1=\{x:Tx=0\},V_2=\{x:h(T)x=0\}$.
Clearly $V_1=\ker{T}$. 
Since $Tx\ne 0$ for any $x\in V_2$ and $x\ne0$, $V_2=\text{Im}{T}$.
