[proofplan]
We prove linear independence by contradiction. Suppose a non-trivial linear combination of the chain vectors vanishes. Applying the nilpotent operator $N$ repeatedly to this relation, we isolate the leading coefficient and show it must be zero. Iterating this argument forces all coefficients to vanish, contradicting the assumption of a non-trivial relation.
[/proofplan]
[step:Suppose a non-trivial linear relation holds among the chain vectors]
Assume for contradiction that there exist scalars $c_0, c_1, \ldots, c_{m-1} \in \mathbb{C}$, not all zero, such that
\begin{align*}
c_0 v + c_1 Nv + c_2 N^2 v + \cdots + c_{m-1} N^{m-1} v &= 0.
\end{align*}
Let $j = \min\{i : c_i \neq 0\}$ be the index of the first non-zero coefficient. Then the relation becomes
\begin{align*}
c_j N^j v + c_{j+1} N^{j+1} v + \cdots + c_{m-1} N^{m-1} v &= 0.
\end{align*}
[guided]
We want to show that the vectors $v, Nv, \ldots, N^{m-1}v$ are linearly independent. The standard approach for proving linear independence is to assume a non-trivial linear combination equals zero and derive a contradiction.
So suppose there exist scalars $c_0, c_1, \ldots, c_{m-1} \in \mathbb{C}$, not all zero, such that
\begin{align*}
c_0 v + c_1 Nv + c_2 N^2 v + \cdots + c_{m-1} N^{m-1} v &= 0.
\end{align*}
Since not all $c_i$ are zero, define $j = \min\{i : c_i \neq 0\}$. All terms with index below $j$ vanish, so the relation simplifies to
\begin{align*}
c_j N^j v + c_{j+1} N^{j+1} v + \cdots + c_{m-1} N^{m-1} v &= 0.
\end{align*}
The key idea is that we can exploit the nilpotency of $N$ to kill higher-order terms while preserving the leading term.
[/guided]
[/step]
[step:Apply $N^{m-1-j}$ to isolate the leading coefficient]
Apply the operator $N^{m-1-j}$ to both sides of the relation. Since $N$ is nilpotent with $N^m v = 0$ (as $v, Nv, \ldots, N^{m-1}v$ is a Jordan chain of length $m$, the chain terminates because $N^m v = 0$), we have for each $i \geq j$:
\begin{align*}
N^{m-1-j}(c_i N^i v) &= c_i N^{m-1-j+i} v.
\end{align*}
When $i > j$, the exponent satisfies $m - 1 - j + i > m - 1 - j + j = m - 1$, so $m - 1 - j + i \geq m$, giving $N^{m-1-j+i} v = 0$. When $i = j$, the exponent is $m - 1 - j + j = m - 1$, giving $c_j N^{m-1} v$.
Therefore, applying $N^{m-1-j}$ yields
\begin{align*}
c_j N^{m-1} v &= 0.
\end{align*}
[guided]
The strategy is to apply a suitable power of $N$ to kill all but the leading term. How do we choose the right power? We want $N^{\ell}(N^i v) = N^{\ell + i} v = 0$ for all $i > j$, while $N^{\ell + j} v \neq 0$. Since the chain has length $m$, we know $N^{m-1} v \neq 0$ (otherwise the chain would have length at most $m - 1$) and $N^m v = 0$.
Setting $\ell = m - 1 - j$, we compute $N^{m-1-j}(c_i N^i v) = c_i N^{m-1-j+i} v$. For each term with $i > j$:
\begin{align*}
m - 1 - j + i &\geq m - 1 - j + (j + 1) = m,
\end{align*}
so $N^{m-1-j+i} v = 0$ since $N^m v = 0$. For the leading term $i = j$:
\begin{align*}
m - 1 - j + j &= m - 1,
\end{align*}
so the leading term survives as $c_j N^{m-1} v$. Applying $N^{m-1-j}$ to the entire relation:
\begin{align*}
c_j N^{m-1} v &= 0.
\end{align*}
[/guided]
[/step]
[step:Conclude that $c_j = 0$, contradicting the choice of $j$]
Since $v, Nv, \ldots, N^{m-1}v$ is a Jordan chain of length $m$, the vector $N^{m-1}v$ is the last non-zero element of the chain, so $N^{m-1}v \neq 0$. From $c_j N^{m-1}v = 0$ and $N^{m-1}v \neq 0$, we conclude $c_j = 0$.
This contradicts the definition of $j$ as the index of the first non-zero coefficient. Therefore no non-trivial linear combination of $v, Nv, \ldots, N^{m-1}v$ can vanish, and the $m$ vectors are linearly independent.
[guided]
We have established that applying $N^{m-1-j}$ to the linear relation kills every term except the leading one, leaving $c_j N^{m-1} v = 0$. The question now is: can $N^{m-1}v$ be zero?
By definition, $v, Nv, \ldots, N^{m-1}v$ is a Jordan chain of length $m$. This means that all $m$ vectors in the chain are non-zero — if $N^{m-1}v$ were zero, then the chain would terminate at $N^{m-2}v$ and have length at most $m-1$, contradicting the assumption that it has length $m$. Therefore
\begin{align*}
N^{m-1}v &\neq 0.
\end{align*}
Now we use a basic property of vector spaces: if $\alpha w = 0$ for a scalar $\alpha$ and a vector $w \neq 0$, then $\alpha = 0$. Applying this to $c_j N^{m-1}v = 0$ with $N^{m-1}v \neq 0$, we conclude
\begin{align*}
c_j &= 0.
\end{align*}
But $j$ was defined as $\min\{i : c_i \neq 0\}$, the index of the first non-zero coefficient. We have just shown $c_j = 0$, which directly contradicts the definition of $j$. This contradiction arose from the assumption that a non-trivial linear combination of the chain vectors equals zero.
Therefore no such non-trivial linear combination exists, and the vectors $v, Nv, \ldots, N^{m-1}v$ are linearly independent.
[/guided]
[/step]
