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For arbitrary square matrix $J$ (diagonalizable or not), can we always find a diagonal matrix $\Lambda$ that $J-\Lambda$ is diagonalizable?

Following is an example that some matrix can fullfill the above requirement. $J=\left(% \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array}% \right)$, $\Lambda=\left(% \begin{array}{cc} -1 & 0 \\ 0 & 0 \\ \end{array}% \right)$. Then $J-\Lambda=\left(% \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ \end{array}% \right)$ is diagonalizable while $J$ is not diagonalizable. (Thanks Marc van Leeuwen for the example.)

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    A nondiagonalizable matrix always has repeated eigenvalues, i.e. multiple roots to its characteristic equation. This is a very fragile situation, and it seems certain that we can find a diagonal perturbation that breaks up the multiple roots into distinct roots. – 2017-01-10
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    A result of Shmuel Friedland ("Matrices with Prescribed Off-Diagonal Elements") shows that it is always possible to construct a complex matrix with a prescribed list of eigenvalues and off-diagonal elements. This implies that if $J$ is complex then by choosing $\Lambda$ appropriately we can make $J - \Lambda$ have any prescribed list of eigenvalues we want and, in particular, we can make the eigenvalues distinct to make $J - \Lambda$ diagonalizable. This is of course much stronger than what the OP asks and works only over the complex numbers. – 2017-01-10
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    A possible avenue for an algebraic proof: the discriminant of the characteristic polynomial of $(J-\Lambda)$ is a polynomial on the entries of $\Lambda$. It suffices to show that this is not the zero polynomial. I wonder if this result holds over general algebraically closed fields, or even finite fields. – 2017-01-10

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Consider the matrix $A_\epsilon = \epsilon J + \operatorname{diag}(1,2,...,n)$.

For $\epsilon>0$ small enough, the Gershgorin discs around the numbers $1,...,n$ are small enough that they do not overlap, in which case the eigenvalues of $J_\epsilon$ are distinct. Then $J+{1 \over \epsilon} \operatorname{diag}(1,2,...,n)$ is diagonalisable.

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    I never seem to think about Gershgorin when it's useful. Good one! – 2017-01-10
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    @Omnomnomnom: I think this is the first time I have explicitly used it. Continuity of eigenvalues is a slightly vague statement that the Gershgorin discs make concrete. I would guess that tiny perturbations should work too, but a proof is not jumping out at me, at least for a small diagonal perturbation. – 2017-01-10