State and prove the Schur Decomposition Theorem.
Use Schur to prove that a square matrix \(A\) has an orthonormal basis of eigenvectors if and only if \(A^H A = AA^H\).
Solution
Part (a).
1. For every \(A\in\mathbb C^{n\times n}\) there is a unitary matrix \(Q\) such that\[ Q^HAQ=T \]is upper triangular.
We argue by induction on \(n\). For \(n=1\) the assertion is immediate. Assume \(n>1\). Since the characteristic polynomial of \(A\) splits over \(\mathbb C\), choose an eigenvalue \(\lambda\) and a unit eigenvector \(q_1\) with \[ Aq_1=\lambda q_1. \] Extend \(q_1\) to an orthonormal basis \(q_1,\ldots,q_n\) of \(\mathbb C^n\), and let \[ Q_1=(q_1\ \cdots\ q_n). \] Then \(Q_1\) is unitary and, because the first column of \(Q_1^HAQ_1\) is \(\lambda e_1\), it has block form \[ Q_1^HAQ_1= \begin{pmatrix} \lambda & *\\ 0 & B \end{pmatrix} \] for some \((n-1)\times(n-1)\) matrix \(B\). By the induction hypothesis there is a unitary \(U\) such that \(U^HBU\) is upper triangular. Set \[ Q=Q_1 \begin{pmatrix} 1&0\\0&U \end{pmatrix}. \] Then \(Q\) is unitary and \[ Q^HAQ= \begin{pmatrix} \lambda & *\\ 0 & U^HBU \end{pmatrix} \] is upper triangular.
2. The diagonal entries of the Schur triangular matrix are the eigenvalues of \(A\), counted with algebraic multiplicity.
Unitary similarity preserves the characteristic polynomial. For an upper-triangular matrix \(T\), the characteristic polynomial is \[ \det(tI-T)=\prod_{j=1}^n(t-t_{jj}), \] so its diagonal entries are precisely its eigenvalues with multiplicity.
Thus 1 and 2 are the Schur Decomposition Theorem.
Part (b).
3. If \(A\) has an orthonormal basis of eigenvectors, then \(A\) is normal:\[ A^HA=AA^H. \]
Let \(Q\) be the unitary matrix whose columns are an orthonormal eigenbasis. Then \[ Q^HAQ=D \] is diagonal, so \[ A=QDQ^H, \qquad A^H=QD^HQ^H. \] Because diagonal matrices commute with their adjoints, \[ A^HA=QD^HDQ^H=QDD^HQ^H=AA^H. \]
4. Every normal upper-triangular complex matrix is diagonal.
Let \(T=(t_{ij})\) be upper triangular and normal. Comparing the \((1,1)\) entries of \(TT^H\) and \(T^HT\) gives \[ \sum_{j=1}^n|t_{1j}|^2=|t_{11}|^2, \] because the first column of an upper-triangular matrix has only the entry \(t_{11}\). Hence \[ t_{1j}=0\qquad(j>1). \] Thus \[ T= \begin{pmatrix} t_{11}&0\\0&T_1 \end{pmatrix}. \] Normality of \(T\) implies normality of \(T_1\). Induction on the size of the matrix now shows that \(T_1\) is diagonal, hence so is \(T\).
5. If \(A\) is normal, then \(A\) has an orthonormal basis of eigenvectors.
By Schur decomposition, choose a unitary \(Q\) such that \[ T=Q^HAQ \] is upper triangular. Normality is preserved by unitary similarity, so \(T\) is normal. By 4, \(T\) is diagonal. Therefore \[ A=QTQ^H \] is unitarily diagonalizable, and the columns of \(Q\) form an orthonormal eigenbasis of \(A\).
6. Consequently, a complex square matrix has an orthonormal basis of eigenvectors if and only if it is normal.
Combine 3 and 5.