Use Schur’s Theorem to prove that every square matrix is arbitrarily close to a diagonalizable matrix.
Show that a square matrix matrix \(A\) has an orthonormal basis of eigenvectors iff \(A^HA=AA^H\) (\(A^*A=AA^*\)).
Solution
1. Every complex square matrix is arbitrarily close, in Frobenius norm, to a diagonalizable matrix.
Let \(A\in M_n(\mathbb C)\) and let \(\varepsilon>0\). By Schur’s theorem there is a unitary matrix \(U\) such that \[ T=U^*AU \] is upper triangular. Write its diagonal entries as \[ \lambda_1,\ldots,\lambda_n. \]
Choose complex numbers \(\delta_1,\ldots,\delta_n\) such that \[ |\delta_j|<\frac{\varepsilon}{\sqrt n} \] for every \(j\), and such that the numbers \[ \lambda_1+\delta_1,\ldots,\lambda_n+\delta_n \] are pairwise distinct. This can be done inductively: after \(\delta_1,\ldots,\delta_{j-1}\) are chosen, only finitely many values of \(\delta_j\) are forbidden by the equalities \[ \lambda_j+\delta_j=\lambda_i+\delta_i\qquad(i<j), \] while the disk \(|\delta_j|<\varepsilon/\sqrt n\) contains infinitely many choices.
Set \[ T_\varepsilon=T+\operatorname{diag}(\delta_1,\ldots,\delta_n) \] and \[ A_\varepsilon=UT_\varepsilon U^*. \] Then \(T_\varepsilon\) is upper triangular with \(n\) distinct diagonal entries, hence has \(n\) distinct eigenvalues and is diagonalizable. Therefore \(A_\varepsilon\), being similar to \(T_\varepsilon\), is diagonalizable.
Finally, the Frobenius norm is unitarily invariant, so \[ \|A_\varepsilon-A\|_F =\|T_\varepsilon-T\|_F =\left(\sum_{j=1}^n|\delta_j|^2\right)^{1/2} <\varepsilon. \] Thus diagonalizable matrices are dense in \(M_n(\mathbb C)\). This proves part (a).
2. If \(A\) has an orthonormal basis of eigenvectors, then\[ A^*A=AA^*. \]
Let \(u_1,\ldots,u_n\) be an orthonormal eigenbasis, with \[ Au_j=\lambda_j u_j. \] Let \[ U=[u_1\ \cdots\ u_n], \qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \] Then \(U\) is unitary and \[ A=UDU^*. \] Hence \[ A^*A =UD^*DU^*, \qquad AA^* =UDD^*U^*. \] Since diagonal matrices commute with their adjoints, \[ D^*D=DD^*, \] so \(A^*A=AA^*\).
3. Every upper-triangular normal matrix is diagonal.
We argue by induction on the size \(n\). The statement is trivial for \(n=1\).
Let \(T=(t_{ij})\in M_n(\mathbb C)\) be upper triangular and normal. Comparing the \((1,1)\) entries of \[ TT^*=T^*T \] gives \[ \sum_{j=1}^n|t_{1j}|^2 = \sum_{j=1}^n|t_{j1}|^2. \] Because \(T\) is upper triangular, \[ t_{j1}=0\qquad(j>1), \] so the right-hand side is \(|t_{11}|^2\). Therefore \[ \sum_{j=2}^n|t_{1j}|^2=0, \] and hence \[ t_{1j}=0\qquad(j>1). \] Thus \[ T= \begin{pmatrix} t_{11}&0\\ 0&B \end{pmatrix} \] with \(B\) upper triangular. Normality of \(T\) implies \[ BB^*=B^*B, \] so \(B\) is normal. By the induction hypothesis, \(B\) is diagonal. Hence \(T\) is diagonal.
4. If \(A^*A=AA^*\), then \(A\) has an orthonormal basis of eigenvectors.
Assume \(A\) is normal. By Schur’s theorem there is a unitary \(U\) such that \[ T=U^*AU \] is upper triangular. Unitary similarity preserves normality, since \[ T^*T =U^*A^*AU, \qquad TT^* =U^*AA^*U. \] Thus \(T\) is upper triangular and normal. By 3, \(T\) is diagonal. Therefore \[ A=UTU^* \] is unitarily diagonalizable, and the columns of \(U\) form an orthonormal basis of eigenvectors of \(A\).
5. Hence a complex square matrix has an orthonormal basis of eigenvectors if and only if it is normal, equivalently\[ A^*A=AA^*. \]
4. This proves part (b).