Making a non-diagonalizable matrix diagonalizable with an small perturbations

Question

For arbitrary non-diagonalizable square matrix $J$, can we always find a arbitrarily small perturbations matrix $\varepsilon A$ that $J+\varepsilon A$ is diagonalizable?

Using Jordan form as following, we can obtain that arbitrarily small pertubations matrix following a certain structure can make a matrix diagonalizable. But can we relax the form of the pertubations matrix?

Give any $J$, let $B$ be the Jordan form. That is, $J=U B U^{-1}$. For a pertubations matrix $\frac{1}{k} U\Delta U^{-1}$ where $\Delta$ is a diagonalizable matrix with different diagonal value, $J+\frac{1}{k} U\Delta U^{-1}$ is diagonalizable.

Can we relax the form of the pertubations matrix?

Answer 1

Any Matrix $A$ has a Jordan normal form. That is by applying a change of basis (call the matrix $S$) to get to a block-diagonal matrix.

Given a Jordan block $J=\begin{pmatrix} \lambda & 1 \\ 0 &\lambda \end{pmatrix}$, it is easy to find an pertubation $E=\begin{pmatrix}0 &0\\0 &\epsilon\end{pmatrix}$, to make $J+E$ diagonalizable (It is a triangular matrix with all different entries on the diagonal).

Use $A+SES^{-1} = S(J+E)S^{-1}$ to see, that the matrix in the middle can be diagonilized with Basis $B$, therefore leading to $SB \sum B^{-1}S^{-1} = (SB)\sum (SB)^{-1}$.

Keep in mind, that the scale of pertubation can be made arbitrary small, since you just want to shove that eigenvalue just a little, little bit.

Keep in mind, that the Jordan normal-form is very instable in numerics.