# Is a square matrix $A$ where $A^3$ is the zero matrix invertible when added to the identity matrix $(I+A)$?

Question: Let $I$ be the $n \times n$ identity matrix and let $O$ be the $n \times n$ zero matrix. Suppose $A$ is a $n \times n$ matrix such that $A^3 = O$. Show that $I + A$ is invertible.

The place that I am stuck at is how do I know about any properties of matrix $A$ to show that it is invertible. I realized that $A$ doesn't necessarily have to be a zero matrix, to begin with.

Additionally, I was wondering if there are any theorems or proofs that would demonstrate that any matrix plus the identity matrix are invertible.

Any help in the right direction would be much appreciated. Cheers.

• Compute $(I+A)(I-A+A^2)$ Commented Feb 23, 2018 at 16:45

You should know the theorem that if $(I+A)x\ne0$ for every $x\ne0$ then $I+A$ is invertible.

So assume that $x\ne0$. You need to show that $(I+A)x\ne0$. If on the other hand $(I+A)x=0$ then $Ax=-x$, and this is impossible because...

• This would be impossible because you would end up with A = -1, if i am understanding that correctly? Commented Feb 23, 2018 at 16:53

Hint:

If $A^3=0$ then $$I=I+A^3=(I+A)(I-A+A^2)$$

• Thanks for the help! Commented Feb 23, 2018 at 17:05

The answers above have already been quite clear. I shall add some motivations (at least for me) how to come up with the inverse.

Now you can pretend that there were a Taylor expansion $$\dfrac{1}{I+A} = I - A + A^2 - A^3 + \cdots .$$ From this, we see that the term $$- A^3 + A^4 + \cdots$$ vanishes. Hence we can guess that the inverse of $$I+A$$ is $$I - A + A^2$$.

Yet this cannot be a correct answer since we haven't dealt with the convergence stuff. However, we can directly verify that $$(I+A)(I - A + A^2) = I.$$ This is what we desired.

So more generally, we can conclude that if $$A^n = 0$$ for some positive integer $$n$$ (such operators are called nilpotent operators), then $$I+A$$ and $$I-A$$ are invertible.

With similar methods, you can try to calculate the inverse of $$I + A + \dfrac{1}{2!} A^2 + \cdots + \dfrac{1}{(n-1)!} A^{n-1}$$ for nilpotent operator $$A$$ with $$A^n = 0$$.

Remark: Such discussions can be carried out in arbitary commutative unital rings.

Remark: Let $$\lambda \in \mathbb{C}$$ (or any appropriate base field), then if $$A$$ is nilpotent, can we discuss the invertibility of $$\lambda I - A$$ or $$I - \lambda A$$? What can we conclude from such calculations?

Hint

We have that

$$(A+I)^3=A^3+3A^2+3A+I=3A^2+3A+I=3A(A+I)+I.$$

In other words

$$(A+I)((A+I)^2-3A)=I.$$

Can you finish?

• Would this be correct then: (I+A)^-1 = I - A - A^2 Commented Feb 23, 2018 at 17:04
• Thank you for the help! Commented Feb 23, 2018 at 17:05
• The last equality says that $A+I$ has inverse and it is $(A+I)^2-3A.$ The same answer as the given by @EmilioNovati.
– mfl
Commented Feb 23, 2018 at 17:06