Linear algebra, help understanding Cayley + Jordan form On page 107 of Evan Chen's Napkin, I don't understand why T is indecomposable means there is an eigenvalue.
On page 110 I don't get how $(X- \lambda )^d$ is the zero map.
 A: 
I don't understand why T is indecomposable means there is an eigenvalue.

Jordan form $J$ of a matrix $A$ is
$$
J = M^{-1}AM=\begin{bmatrix}
       J_1 &  \\ 
        & . \\
        & & .\\
        & & &J_s
    \end{bmatrix}\qquad(1)
$$
where each block $J_i$ is a Jordan block, given by
$$
J_i = \begin{bmatrix}
       \lambda_i & 1 \\ 
        & \lambda_i & . \\
        & & . &1\\
        & & &\lambda_i
    \end{bmatrix}\qquad(2)
$$
This Jordan block represents a single eigenvalue $\lambda_i$ of the matrix $A$
So $T$ is indecomposable means that $T$ is of the form $J_i$, in this case $T:V_1 \to V_1 $ , $V_1 = [J_1]$ 
and since $J_1$ represents a single eigenvalue of $A$, there is an eigenvalue.

I don't get how $(X−λ)^d$ is the zero map.

From Pg.110 of the pdf you linked,
$$
p_T(X) = (X−λ_1I)^{d_1}(X−λ_2I)^{d_2}...(X−λ_mI)^{d_m}
$$
Writing each $X$ in its Jordan form, $X = MJM^{-1}$,
$$
p_T(J) = (MJM^{-1}−λ_1MM^{-1})^{d_1}(MJM^{-1}−λ_2MM^{-1})^{d_2}...(MJM^{-1}−λ_mMM^{-1})^{d_m}
$$
$$
p_T(J) = \bigl(M(J-λ_1I)M^{-1}\bigr)^{d_1}\bigl(M(J-λ_2I)M^{-1}\bigr)^{d_2}...\bigl(M(J-λ_mI)M^{-1}\bigr)^{d_m}
$$
Now concentrating on each $J-λ_iI$, from eq(1) and eq(2), this produces an upper triangular block matrix with $0$s on diagonal. For each term in the product, there are $0$s on corresponding diagonals and their product will result in a Zero matrix.
Edit:
Explanation on how T acts independently on each subspace of Jordan block (pg. 105 of pdf linked).
Let's say for a $T:V \to V$ such that T has a complete set of independent eigenvectors. Taking the example given in the pdf:
$$
T = \begin{bmatrix}
\ 3\\
& 7\\
& & 2\\
& & & 2\\
& & & & 5\\
& & & & & 3
\end{bmatrix}
$$
For this matrix, the eigenvectors are $3e_1, 7e_2, 2e_3, 2e_4, 5e_5, 3e_6$ and each eigenvector has an eigenspace.
When $T$ acts on each of the eigenspace (of each eigenvector), to produce $Tv = \lambda v$, these are also independent since they are scalar multiples of independent eigenvectors.
Now suppose that $T$ cannot be diagonalized and a complete set of independent eigenvectors cannot be extracted. In that case, lets say $\lambda_1$ is a repeated eigenvalue and $\lambda_2$ is a single eigenvalue, for which $\lambda_1$ has only one independent eigenvector $v_1$ and $\lambda_2$ has eigenvector $v_2$.
Now in order to produce an upper triangular matrix in Jordan form, we choose the basis as $v_1$, $v_1'$ and $v_2$ such that, $Tv_1 = \lambda_1 v_1$, $Tv_2 = \lambda_2 v_2$ and finally $Tv_1' = v_1 + \lambda_1 v_1'$. i.e. check that 
$$
\begin{bmatrix}
\ v_1'\\
v_1\\
v_2
\end{bmatrix}
T = J
\begin{bmatrix}
\ v_1'\\
v_1\\
v_2
\end{bmatrix}
$$
where
$$
J = \begin{bmatrix}
\ \lambda_1 & 1 \\
& \lambda_1 \\
& & \lambda_2
\end{bmatrix}
$$
Multiply yourself and verify.
This gives us the Jordan form, $VT = JV$ $\Rightarrow$ $VAV^{-1} = J$
So to conclude, $T$ applies painlessly on $v_2$ to produce a diagonal element. In order to transform T to a form which is as diagonal as possible, we had to take a basis vector $v_1'$ along with $v_1$ which produces a Jordan block with diagonal elements and an extra '1' above the diagonal. Applying T on this basis (which is actually a subspace of $v_1$), produced an independent Jordan block. This when translated to n dimensions, shows that T applies independently on each subspace of $v$ to give its constituent Jordan blocks.
