📂Matrix Algebra

Every matrix is the limit of a sequence of diagonalizable matrices

Theorem¹

For every matrix $A \in M_{n \times n}(\mathbb{C})$, there exists a sequence of diagonalizable matrices that converges to $A$ (converges).

Proof

If $A$ is diagonalizable, the claim is trivial (see), so assume $A$ is not diagonalizable. Since every matrix is similar to an upper-triangular matrix (see), there exist an invertible matrix $S$ and an upper-triangular matrix $T$ satisfying

$$ A = S^{-1}TS, \qquad T = \begin{bmatrix} \lambda_{1} & * & \cdots & * \\ 0 & \lambda_{2} & \cdots & * \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{bmatrix} $$

Here $\lambda_{i}$ denotes an eigenvalue of $A$ counted with multiplicity (since $A$ is not diagonalizable). Now define the diagonal matrix $D_{k} = \diag(0, \frac{1}{k}, \frac{2}{k}, \cdots, \frac{n-1}{k})$. If we set $T_{k} = T+D_{k}$, then $T_{k}$ has $n$ distinct diagonal entries (= eigenvalues) and hence is diagonalizable. If $A_{k} = S^{-1} T_{k} S$, then $A_{k}$ is diagonalizable and converges to $A$.

$$ \begin{align*} \| A_{k} - A \| &= \| S^{-1} T_{k} S - S^{-1}TS \| \\ &= \| S^{-1}(T_{k}-T)S \| \\ &\le \| S^{-1} \| \| T_{k}-T \| \| S \| \to 0 \text{ as } k \to \infty \end{align*} $$

The inequality follows from the property of matrix norms $\| A B \| \le \| A \| \| B \|$ (see).

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