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 $$ (A+B)^{-1} = B^{-1}(I+B^{-1}A)^{-1}. $$

  • 2
    $\begingroup$ 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 $i = k - 1$. $\endgroup$ Oct 14 '14 at 16:04
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    $\begingroup$ @MichaelJoyce: sure, but it's not wrong to go up to $n$... $\endgroup$
    – anderstood
    Oct 14 '14 at 22:09
  • $\begingroup$ @anderstood of course, but then you need to know/prove $k\le n$. Changed it to $k-1$ to be on the 'safe' side. $\endgroup$
    – daw
    Oct 15 '14 at 6:17
  • $\begingroup$ @anderstood: You're absolutely right. My nitpick was a (very, very minor) aesthetic one, not a critique of the actual content. $\endgroup$ 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?


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