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No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $A=\begin{bmatrix}a/3 & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}1\\1\\1\end{bmatrix}$ to $\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}$ (for instance, a Householder reflector).

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $A=\begin{bmatrix}3a & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}\sqrt {3}\\0\\0\end{bmatrix}$ to $\begin{bmatrix}1\\1\\1\end{bmatrix}$ (for instance, a Householder reflector).

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $\begin{bmatrix}a/3 & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}1\\1\\1\end{bmatrix}$ to $\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}$ (for instance, a Householder reflector).

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $A=\begin{bmatrix}a/3 & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}1\\1\\1\end{bmatrix}$ to $\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}$ (for instance, a Householder reflector).

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

No, there are no such bounds. Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $\begin{bmatrix}a & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a matrix that maps $\begin{bmatrix}1\\0\\0\end{bmatrix}$ to $\begin{bmatrix}1\\1\\1\end{bmatrix}$ (an orthogonal one, for instance). Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $\begin{bmatrix}a/3 & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}1\\1\\1\end{bmatrix}$ to $\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}$ (for instance, a Householder reflector). 

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.