Let $A$ and $Q$ be linear operators on finite dimensional vector space. We know that Q is nilpotent and it commutes with A. Prove that operator $(A + Q)^{-1}$ is invertible if and only if $A$ is invertible.

What I have done so far:

Let suppose A is invertible, than:

$$(A + Q)^{-1} = \frac{1}{A+Q} = \frac{A^{-1}}{1+A^{-1}Q} = A^{-1} \Bigl(\frac{1}{1-(-A^{-1}Q)}\Bigr)$$

Which I re-wrote as:

$$A^{-1}\bigl(1 + (-A^{-1}Q) + (-1A^{-1}Q)^2...\bigr)$$

Now since Q is nilpotent, it means there exist such $m$, such that $Q^m = 0$.

Which means, that from some term onward, all terms are zero so I have: $$A^{-1}\bigl(1 -A^{-1}Q + A^{-2}Q^2... A^{-m}Q^m(-1)^m \bigr)$$

Which shows $(A+Q)^{-1}$ is invertible.

My question is, how do I prove ekvivalence in other direction?


1 Answer 1


If $A + Q$ is invertible then $-Q$ is nilpotent and commutes with $A + Q$ so $(A + Q) + (-Q)$ is invertible.

  • $\begingroup$ I see why $A + Q$ commutes with $-Q$, but could you explain why does it follow that $(A+Q) + (-Q) $is invertible. $\endgroup$ Commented Jan 8, 2020 at 21:30
  • 1
    $\begingroup$ Because you proved that if $Q$ commutes with $A$ and $A$ is invertible then $A + Q$ is invertible. Well here $-Q$ commutes with $A+Q$ and $A + Q$ is invertible. @UrošKosmač $\endgroup$
    – Sera Gunn
    Commented Jan 8, 2020 at 22:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .