I was reading proof of existence of Schur decomposition.
I understand everything except one thing.
Why submatrices like $A_2$ have same eigen value with main matrix $A$ like $\lambda_2$ ?
A triangular matrix has the property that its diagonal entries are equal to its eigenvalues. Moreover, two similar matrices have the same eigenvalues. Therefore,
the Schur decomposition $T = Q^{*}AQ$
allows us to read the eigenvalues of A on the main diagonal of $T$, which is upper triangular and similar to A.
Best Answer
One way to look at it is that, we can assume $A_2$ has some eigenvalue $\mu_2$ and get on with the proof. The proof will go through even if we do not assume $\mu_2$ as an eigenvalue of $A$ but an eigenvalue of $A_2$. Finally after Schur decomposition is complete, $\mu_2$ will be one of the diagonal entries and it will be an eigenvalue of $A$ by virtue of the fact that similarity transformation preserves the eigenvalues and $\mu_2$ is an eigenvalue of $T$. This is one mathematically consistent way of understanding this.
The reduced matrix is $T_1 = \begin{bmatrix} \lambda_1 & *\\ 0 & A_2 \end{bmatrix}$. If lets say $v_2$ is an eigen vector of reduced matrix $T_1$ with eigenvalue of $\lambda_2$ then $T_1 v_2 = \lambda_2 v_2$ and hence $[0,A_2]v_2 = \lambda_2 v_{2,2:end}$ and hence $v_{2,2:end}$ is an eigenvector of $A_2$ with eigenvalue $\lambda_2$. Note that since $\lambda_2$ is an eigenvalue of $T_1$, it is also an eigenvalue of $A$. Hence eigenvalues of $T_1$ (and hence eigenvalues of $A$) are also eigen values of $A_2$ except $\lambda_1$.