State and prove the Schur Decomposition Theorem.
Solution
1. Schur Decomposition Theorem. For every matrix \(A\in M_n(\mathbb C)\), there exists a unitary matrix \(U\in U(n)\) such that\[ U^*AU=T \]is upper triangular. The diagonal entries of \(T\) are the eigenvalues of \(A\), counted with algebraic multiplicity.
We prove the triangularization statement by induction on \(n\).
2. The theorem holds for \(n=1\).
Every \(1\times1\) matrix is already upper triangular, and the identity matrix is unitary.
3. Assume \(n>1\). Since the characteristic polynomial of \(A\) splits over \(\mathbb C\), choose an eigenvalue \(\lambda\) and a unit eigenvector \(v_1\) such that\[ Av_1=\lambda v_1. \]Extend \(v_1\) to an orthonormal basis\[ v_1,v_2,\ldots,v_n \]of \(\mathbb C^n\), and let\[ U_1=[v_1\ v_2\ \cdots\ v_n]. \]Then \(U_1\) is unitary and\[ U_1^*AU_1= \begin{pmatrix} \lambda & *\\ 0 & B \end{pmatrix} \]for some \(B\in M_{n-1}(\mathbb C)\).
The columns of \(U_1\) form an orthonormal basis, so \(U_1\) is unitary. The first column of \(U_1^*AU_1\) is the coordinate vector of \(Av_1=\lambda v_1\) in this basis, namely \[ (\lambda,0,\ldots,0)^T. \] Thus every entry below the \((1,1)\) entry in the first column is zero, giving the displayed block form.
4. By the induction hypothesis, there exists a unitary matrix \(W\in U(n-1)\) such that\[ W^*BW=T_0 \]is upper triangular.
The matrix \(B\) is an arbitrary complex \((n-1)\times(n-1)\) matrix, so the induction hypothesis applies to it.
5. Put\[ D= \begin{pmatrix} 1&0\\ 0&W \end{pmatrix} \qquad\text{and}\qquad U=U_1D. \]Then \(U\) is unitary and\[ U^*AU = \begin{pmatrix} \lambda&*\\ 0&T_0 \end{pmatrix}, \]which is upper triangular.
Both \(U_1\) and \(D\) are unitary, hence so is their product \(U\). Moreover \[ U^*AU =D^*U_1^*AU_1D = \begin{pmatrix} 1&0\\ 0&W^* \end{pmatrix} \begin{pmatrix} \lambda&*\\ 0&B \end{pmatrix} \begin{pmatrix} 1&0\\ 0&W \end{pmatrix} = \begin{pmatrix} \lambda&*\\ 0&W^*BW \end{pmatrix}. \] By 4, the lower-right block is upper triangular, so the whole matrix is upper triangular.
6. The diagonal entries of \(T=U^*AU\) are the eigenvalues of \(A\), counted with algebraic multiplicity.
Because \(T\) is similar to \(A\), they have the same characteristic polynomial. Since \(T\) is upper triangular, \[ \det(tI-T)=\prod_{j=1}^n(t-t_{jj}). \] Thus the roots of the characteristic polynomial, with multiplicity, are exactly the diagonal entries \(t_{11},\ldots,t_{nn}\).