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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?

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    Are you asking about permutations or pertubations? The first is matrix with a single $1$ in every row and collumn, the latter is an arbitrary matrix with small entries. – 2017-02-02
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    What is "a **small**" permutation matrix? – 2017-02-02
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    I have updated the question. permutations to pertubations and small to arbitrarily small. Thanks. – 2017-02-02
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    That depends on the field. – 2017-02-02
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    Is $A$ chosen after epsilon? – 2017-02-02
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    @user1551 I think that most users in this site are implicitely using fields $\mathbb{R}$ or $\mathbb{C}.$ – 2017-02-02
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    I am trying to find all the possible forms of $A$. – 2017-02-02
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    I (coming from engineering) rarely see matrices over other fields. I therefore assume, if it is not stated otherwise, that we are in those fields. Since there is, where most linear algebra is taught. – 2017-02-02
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    @JeanMarie Yes, but the answer still depends on the field -- it's "yes" if the field is complex, and "no" if the field is real. – 2017-02-02
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    @Sunnydreamrain Don't you think there is some "perturbation" in your "pertubation" ? :) – 2017-02-02
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    @user1551 I ignored! (i thought you where speaking of very special fields) Why isn't it possible for real entries ? – 2017-02-02
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    @JeanMarie Eigenvalues vary continuously with matrix entries. When the perturbation is small, the perturbed value of any non-real eigenvalue is still non-real. So, if a matrix is not diagonalisable over $\mathbb R$ owing to the presence of non-real eigenvalues, the perturbed matrix is still not diagonalisable over $\mathbb R$. – 2017-02-02
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    @user1551 I see clearly now. Thank you very much – 2017-02-02
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    @user1551 In "Laguerre's Method Applied to the Matrix Eigenvalue Problem" (the second page, section 2, The Condition of an Eigenvalue.) It shows that Eigenvalue can be ill-conditioned, meaning a small perturbation to the matrix can introduce a very large change to the Eigenvalue. Is this contradictory with your comment "Eigenvalues vary continuously with matrix entries". http://www.ams.org/journals/mcom/1964-18-087/S0025-5718-1964-0165668-2/S0025-5718-1964-0165668-2.pdf – 2017-02-02
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    @Sunnydreamrain Sensitivity and continuity are two things. There is no contradiction. – 2017-02-03
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    @user1551 I am not very familiar with this. But what does the following line in your comment mean? "When the perturbation is small, the perturbed value of any non-real eigenvalue is still non-real." – 2017-02-03
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    @Sunnydreamrain Continuity means that the perturbed eigenvalues can be made *arbitrarily* close to their original values if the perturbations to the matrix entries are sufficiently small. If the original eigenvalues are non-real, any numbers that are close to them are non-real too. – 2017-02-03

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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.

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    Hi, this is only for the complex field since Jordan form is defined in the complex field. Is there anyway to show it in the real field? – 2018-03-22