Motivation Behind the Key Step in Schur Decomposition Proof I often find myself forgetting how to prove that every square matrix having a Schur factorization because I never really understood the motivation behind the steps, I only memorized how to do it. I would like to rectify this. Here's the proof I am familiar with:

We proceed by induction. Suppose every $n \times n$ matrix has a Schur decomposition and take an $(n+1) \times (n+1)$ square matrix $A$ and write
$$A = \begin{bmatrix} a & v^*\\
u & B \end{bmatrix}$$
where $a \in \mathbb{C}$, $u, b$ are vectors, and $B$ is $n \times n$. Let $v_1$ be a normalized eigenvector of $A$ with corresponding eigenvalue $\lambda$ and extend $v_1$ to an orthonormal basis $v_1, ..., v_{n+1}$ of $\mathbb{C}^{n+1}$. Define the matrix
$$V = \begin{bmatrix}v_1 \ \  v_2 \ \ \cdots \ \  v_{n+1} \end{bmatrix} = \begin{bmatrix} v_1 \ \  V_2\end{bmatrix}.$$
We have
$$VAV^* = \begin{bmatrix}
v_1^* \\ V_2^*\end{bmatrix}
A
\begin{bmatrix}
v_1 \ \ V_2
\end{bmatrix}
 = \begin{bmatrix}
\lambda & v_1^*AV_2\\
\lambda V_2^*v_1 & V_2^*AV_2
\end{bmatrix}
= \begin{bmatrix}
\lambda & w^*\\
0 & C
\end{bmatrix}$$
where $w, C$ are defined in the obvious way. Now $C$ is $n \times n$, so it has a Schur decomposition $QUQ^*$. Thus
$$VAV^* = \begin{bmatrix} 1 & 0^T\\
0 & Q \end{bmatrix}
\begin{bmatrix}
\lambda & w^*\\
0 & U
\end{bmatrix}
\begin{bmatrix}
1 & 0^T\\
0 & Q^*
\end{bmatrix}
$$
The middle matrix on the right is upper triangular, so solving for $A$ above shows a Schur Decomposition of $A$.

My Analysis of the Proof:
The proof starts by induction which I can accept, as that is a standard technique in Linear Algebra. Then it's natural (but wrong) to write $B$ as its Schur factorization and try to factor out a unitary matrix $K$ on the left and its adjoint $K^*$ on the right; but this doesn't quite work. Instead, we have to conjugate $A$ by a strange unitary matrix $V$ (in which only the first column really matters, not the others) and only THEN the above trick works.
It seems like the key step is to multiply $A$ by this strange matrix $V$. What is the motivation behind this matrix? How could we have known that conjugating $A$ by $V$ will allow us to factor $VAV^*$ in a useful way?
Thank you very much.
 A: I'm using your notation.
Let $F=\mathbb{C}v_1$. Then $\mathbb{C}^{n+1}=F\oplus F^\perp$, where $F^\perp$ has dimension $n$.
So, if you glue $v_1$ and an orthonormal basis of $F^\perp$, you will get an orthonormal basis $\mathcal{B}$ of $\mathbb{C}^{n+1}$. Now the main point is that $Av_1=\lambda v_1$
So if we call $u$ the endomorphism represented by $A$ in the canonical basis, we have $u(v_1)=\lambda v_1$.
So , by definition of a representative matrix, $Mat(u; B)=\begin{pmatrix}\lambda & w^* \cr 0 & C\end{pmatrix}$ (since  $u(v_1)=\lambda v_1+0 v_2+\cdots+0v_{n+1}$).
If $V$ is the base change matrix, then $V$ is orthogonal (since it's a base change matrix between two orthonormal bases), and the base change formula tells you that $V^{-1}AV=V^*AV  =\begin{pmatrix}\lambda & w^* \cr 0 & C\end{pmatrix}$ (The fact that you have $V^*$ on the right is strange, btw).
All in all, as runway44 said, this is just a base change matrix, plus the following general principle of linear algebra:
let $E$ be a $K$-vector space of finite dimension, let $u$ be an endomorphism of $E$. Assume that $E=F\oplus G,$ with $u(F)\subset F$. Glue a basis $B_1$ of $F$ and a basis $B_2$ of $G$ to get a basis $B$ of $E$. Then $Mat(u; B)=\begin{pmatrix}A& A' \cr 0 & C\end{pmatrix}$.
