Proving or disproving matrix $A+B$ is invertible

Given $$A, B \in M_n (\Bbb F)$$, where $$A$$ is $$k$$-nilpotent and $$B$$ is invertible, is $$A+B$$ also invertible?

I was having trouble on how to prove this, and then I thought maybe this statement is incorrect, but couldn't find a counter example. Perhaps someone can assist?

This is not true in general. Take $$A = \begin{pmatrix} 0 &1 \\ 0 & 0 \end{pmatrix}, \quad B= \begin{pmatrix} 0 &-1 \\ 1& 0 \end{pmatrix}.$$ However, $I+A$ is always invertible for nilpotent $A$. The same holds for $A+B$ with invertible $B$ if $A$ and $B$ commute: If $A$ and $B$ commute, then $A$ and $B^{-1}$ commute, which implies that $B^{-1}A$ is nilpotent, moreover $I-(-B^{-1}A)$ is invertible with $$(I-(-B^{-1}A))^{-1} = \sum_{i=0}^{k-1} (-B^{-1}A)^{i}$$ which implies $$(A+B)^{-1} = B^{-1}(I+B^{-1}A)^{-1}.$$

• And now you have a second proof that $AB \neq BA$ for your specific example! ;) Small nitpick: if we're assuming $A$ is $k$-nilpotent, then the sum in your formula needs only to run from $i = 0$ to $i = k - 1$. Oct 14 '14 at 16:04
• @MichaelJoyce: sure, but it's not wrong to go up to $n$... Oct 14 '14 at 22:09
• @anderstood of course, but then you need to know/prove $k\le n$. Changed it to $k-1$ to be on the 'safe' side.
– daw
Oct 15 '14 at 6:17
• @anderstood: You're absolutely right. My nitpick was a (very, very minor) aesthetic one, not a critique of the actual content. Oct 15 '14 at 13:05

A counterexample for $n=2$ by taking

$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$$

Hint: Consider the case $k = 2$ with $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Can you find a matrix $B$ with linearly independent columns such that the columns of $A + B$ become linearly dependent?