Let $K$ be a field s.t. $charac(K)\not= 2$. Let $A\in\mathcal{M}\_n(K)$ be such that $rank(A)\geq n/2$. Can one find a matrix $B$ such that $B$ is similar to $A$ and $A+B$ is invertible ?



Invertible matrix
Initially my assumption was that the question was primarily about finite fields $K$, but after reading the comments it appears to me that there is some confusion about the infinite case as well. So I'll prove the claim for $|K|=\infty$. I would guess the assertion is also true if $K$ is finite, but I don't know how to show it.


If one assumes $K$ to be algebraically closed, then it's very easy to see that one can find an appropriate $B$: Let $\lambda\_1,\ldots,\lambda\_n$ be the eigenvalues of $A$ (eigenvalues with multiplicity $m$ are supposed to appear $m$ times in that list), then for any permutation $\sigma\in S\_n$ we can conjugate $A$ to a matrix of the form
$$
 \textrm{diag}(\lambda\_{\sigma(1)}, \ldots, \lambda\_{\sigma(n)}) + N
$$
where $N$ is a strictly upper triangular matrix.
*(This should be clear but I'll give a short explantion anyhow: View $K^n$ as a $K[x]$ module where $x$ acts as $A$. Then finding a basis that conjugates $A$ into the form given above corresponds to finding a filtration $0= V\_0 < V\_1 <\ldots< V\_n= K^n$ of the $K[x]$-module $K^n$ such that the quotient $V\_i/V\_{i-1}$ is isomorphic to the simple $K[x]$-module $K[x]/(x-\lambda\_{\sigma(i)})$. But by the structure theorem on f. g. modules over PID's, $K^n$ is a direct sum of modules each having just one isomorphism type of simple modules as composition factors. So choosing such a filtration is no problem: in each step just pick a maximal submodule in the desired isotypic direct summand and leave the other summands unchanged.)*


We may assume that $A$ is of the above shape with $\sigma=\textrm{id}$. So just choose a permutation $\sigma$ such that $\lambda\_i+\lambda\_{\sigma(i)} \neq 0$ for all $i$ (this can be done: just choose $\sigma$ to be the transposition that swaps each $\lambda\_i$ which is zero with some $\lambda\_i$ which is non-zero and fixes the rest; the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues; also $char(K)\neq 2$ is essential: it guarantees $\lambda\_i\+\lambda\_i\neq 0$ whenever $\lambda\_i\neq 0$). Then $A+B$ can becomes
$$
\textrm{diag}(\lambda\_1+\lambda\_{\sigma(1)}, \ldots, \lambda\_n+\lambda\_{\sigma(n)}) + \textrm{something strictly upper triangular}
$$
and you are done.


As for the case of a non-algebraically closed (but still infinite) $K$: Note that the algebraically closed case implies that the rational function
$$
 det(A+TAT^{-1})=\frac{det(AT+TA)}{det(T)} \in K(T\_{ij})
$$
is non-zero. But $GL(n,K)$ is Zariski dense in $GL(n,\bar K)$ (see [here](https://mathoverflow.net/questions/56192/density-question-in-algebraic-group)), and so the above rational funktion must be non-vanishing on $GL(n,K)$, which shows there is a $T\in GL(n,K)$ such that $det(A+TAT^{-1})\neq 0$. 


