Triangularization of Matrices
Definition
If a matrix $A \in M_{n \times n}(\mathbb{C})$ is similar to a triangular matrix, it is called triangularizable.
$$ A = P^{-1} U P $$
Here, $U$ is, without loss of generality, assumed to be an upper triangular matrix.
Theorem
Every square matrix is triangularizable. In other words, it is similar to an upper triangular matrix.
Remark
In particular, a nilpotent matrix is similar to an upper triangular matrix whose diagonal entries are $0$.
Proof
Let $A \in M_{n \times n}(\mathbb{C})$. Since every matrix has at least one eigenvalue, let $\lambda_{1}$ be an eigenvalue of $A$ and $v_{1}$ a corresponding eigenvector. By Gram–Schmidt orthogonalization we can obtain an orthonormal basis $\left\{ v_{1}, v_{1,2}, \cdots, v_{1,n} \right\}$. Let the matrix having these as column vectors be $Q_{1}$.
$$ Q_{1} = \begin{bmatrix} \overset{|}{\underset{|}{v_{1}}} & \overset{|}{\underset{|}{v_{1,2}}} & \cdots & \overset{|}{\underset{|}{v_{1,n}}} \end{bmatrix} $$
Then $Q_{1}$ is a unitary matrix, and there exist $\mathbf{y} \in \mathbb{C}^{n}$ and $B \in M_{n \times n}(\mathbb{C})$ such that the following holds.
$$ Q_{1}^{\ast} A Q_{1} = \begin{bmatrix} \lambda_{1} & \mathbf{a}^{\mathsf{T}} \\ \mathbf{0} & B \end{bmatrix} $$
By the same reasoning, there exists an eigenvalue $\lambda_{2}$ of $B$, and we can obtain a unitary matrix $U$ formed from an orthonormal basis that includes a corresponding eigenvector. Then we obtain:
$$ \begin{align*} \begin{bmatrix} 1 & \mathbf{0}^{\mathsf{T}} \\ \mathbf{0} & U^{\ast} \end{bmatrix} Q_{1}^{\ast} A Q_{1} \begin{bmatrix} 1 & \mathbf{0}^{\mathsf{T}} \\ \mathbf{0} & U \end{bmatrix} &= \begin{bmatrix} 1 & \mathbf{0}^{\mathsf{T}} \\ \mathbf{0} & U^{\ast} \end{bmatrix} \begin{bmatrix} \lambda & \mathbf{a}^{\mathsf{T}} \\ \mathbf{0} & B \end{bmatrix} \begin{bmatrix} 1 & \mathbf{0}^{\mathsf{T}} \\ \mathbf{0} & U \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{1} & \mathbf{a}^{\mathsf{T}} \\ \mathbf{0} & U^{\ast} B U \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{1} & a_{12} & \mathbf{b}^{\mathsf{T}} \\ 0 & \lambda_{2} & \mathbf{c}^{\mathsf{T}} \\ \mathbf{0} & \mathbf{0} & C \end{bmatrix} \end{align*} $$
If we denote the unitary matrices constructed in this manner by $\begin{bmatrix} 1 & & & \\ & 1 & & \\ & & \ddots & \\ & & & U_{i} \end{bmatrix}$ and call their product $Q_{i}$, then $Q = Q_{1} \cdots Q_{n}$ is also a unitary matrix. The following holds, and thus $A$ is similar to an upper triangular matrix.
$$ Q^{\ast}AQ = \begin{bmatrix} \lambda_{1} & a_{12} & \cdots & a_{1n} \\ 0 & \lambda_{2} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{bmatrix} $$
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