- Prove that the set of diagonalizable \(N \times N\) matrices over \(\mathbb{C}\) is dense in \(M_N(\mathbb{C})\) in the standard Euclidean/metric topology.
- Prove that over any algebraically closed field \(K\), the set of diagonalizable matrices is Zariski dense in \(M_N(K) \cong \mathbb{A}^{N^2}\).
Solution
Let \(A\in M_N(K)\). If the characteristic polynomial \(\chi_A(t)\) has \(N\) distinct roots, then \(A\) is diagonalizable. The discriminant \[ \Delta(A)=\operatorname{Disc}(\chi_A) \] is a polynomial in the \(N^2\) entries of \(A\), and \[ \Delta(A)\ne0 \] exactly when \(\chi_A\) has distinct roots.
This polynomial is not identically zero. Indeed, over any algebraically closed field \(K\) one can choose \(N\) distinct scalars \(a_1,\dots,a_N\in K\); then for \[ D=\operatorname{diag}(a_1,\dots,a_N) \] we have \[ \Delta(D)=\prod_{i<j}(a_i-a_j)^2\ne0. \] Thus the principal open set \[ U_\Delta=\{A:\Delta(A)\ne0\} \] is nonempty and consists of diagonalizable matrices. Since affine space \(M_N(K)\cong\mathbb A^{N^2}\) is irreducible, every nonempty Zariski-open subset is dense. Hence diagonalizable matrices are Zariski dense in \(M_N(K)\).
For the ordinary topology over \(\mathbb C\), let \(A=UTU^*\) be a Schur triangularization, with diagonal entries \(\lambda_1,\dots,\lambda_N\). Given \(\varepsilon>0\), choose \(\delta_1,\dots,\delta_N\) with \[ |\delta_i|<\varepsilon \] such that the numbers \(\lambda_i+\delta_i\) are pairwise distinct, and set \[ T_\varepsilon=T+\operatorname{diag}(\delta_1,\dots,\delta_N), \qquad A_\varepsilon=UT_\varepsilon U^*. \] Then \(A_\varepsilon\) has distinct eigenvalues and is therefore diagonalizable, while \[ \|A_\varepsilon-A\|=\max_i|\delta_i|<\varepsilon \] for the operator norm. Thus diagonalizable complex matrices are Euclidean-dense as well.