8
$\begingroup$

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?

$\endgroup$
0
10
$\begingroup$

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

$\endgroup$
4
  • 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
  • 1
    $\begingroup$ @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
9
$\begingroup$

A counterexample for $n=2$ by taking

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

$\endgroup$
5
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.