A Nilpotent Matrix is Similar to a Strictly Upper Triangular Matrix
Theorem
Let the matrix $A \in M_{n \times n}(\mathbb{C})$ be nilpotent. Then it is similar to an upper-triangular matrix.
Proof
Assume $A$ satisfies $A^{k} = O$. Then, by the monotonicity of kernels, there exists the following flag.
$$ \left\{ \mathbf{0} \right\} \lneq \ker A \lneq \ker A^{2} \lneq \cdots \lneq \ker A^{k-1} \lneq \ker A^{k} = \mathbb{C}^{n} $$
$$ \dim (\ker A) \lt \dim (\ker A^{2}) \lt \cdots \lt \dim (\ker A^{k-1}) \lt \dim (\ker A^{k}) = n $$
Denote a basis of $\ker A$ by $B_{1}$. Let $B_{2}$ be a basis of $\ker A^{2}$ obtained by extending this basis. In the same way, denote the bases of $\ker A^{i}$ obtained analogously by $B_{i}$. Define the set $S_{i}$ as follows.
$$ S_{i} = B_{i} \setminus B_{i-1} $$
That is, $S_{i}$ is the set of basis vectors newly added in $\ker A^{i}$. If a basis vector is $v \in S_{i}$, then by definition it satisfies
$$ v \in \ker A^{i}, \quad Av \in \ker A^{i-1}, \\ v \notin \ker A^{i-1} $$
But since $\ker A^{i-1} = \span (B_{i-1}) = \span (S_{1} \cup \cdots \cup S_{i-1})$, $Av$ can be expressed as a linear combination of vectors in $S_{1} \cup \dots \cup S_{i-1}$. In other words, if we consider the coordinate vectors with respect to the choice $B_{k}$, the components (coordinates) of $Av$ are non-$0$ only up to the $N_{i-1} = |B_{i-1}|$-th entry, and thereafter all entries are $0$. Without loss of generality(일반성을 잃지않고), suppose all vectors in $B_{k}$ have norm $1$; then the matrix $Q$ whose columns are these vectors is a unitary matrix, and $Q^{\ast}AQ$ is an upper-triangular matrix.
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