# Showing that a matrix is invertible and finding its inverse

I'm incredibly rusty at linear algebra, and in preparation for my course I've been doing some review questions. I've been staring at this one for a half hour and still don't know how to approach it:

"Let A be a square matrix such that $A^3 = 0$. Show that the matrix $I + A + 2A^2$ is invertible and find its inverse."

I'm pretty sure I need to find a relationship between $A^3$ and $I + A + 2A^2$, but I'm not sure how. A matrix is invertible if the determinant is nonzero, and I know how to find the inverse of a matrix, but since this is a more theoretical question I'm not entirely certain how to approach it. Any hints would be much-appreciated :)

We start by an informal deduction: by the Taylor's expansion (around 0) $$(1+x+2x^2)^{-1}=1-x-x^2+(\text{terms with order 3 or above}).$$ We would like to substitute $A$ into $x$. But with $A^3=0$, all the terms with power $3$ or above vanish. This suggests $$(I+A+2A^2)^{-1}=1-A-A^2.$$ Now the solution is made rigorous by direction verification: $$(I+A+2A^2)(I-A-A^2)=I-3A^3-2A^4=I-0-0=I.$$ While perhaps not elegant, this approach is mechanical so it can be applied to similar problems. For example, with $I+a A+b A^2$, we have $$(1+a x+bx^2)^{-1}=1-ax+(a^2-b)x^2+(\text{terms with order 3 or above})\\ \implies (I+aA+bA^2)^{-1}=I-aA+(a^2-b)A^2.$$ Again, rigor will be supplied by verifying the solution via direct multiplication.
• I sort of understand where you're coming from here. Essentially, you're using the idea that $A A^-1 = I$. I don't understand exactly how you found the inverse of that matrix. Is there a different way to do that? To my memory, I've never used Taylor's expansion to solve any problems in the course, so I wonder if there's a simpler solution. – Alex Sep 14 '14 at 0:46
• No, my essential idea is not $AA^{-1}=I$. Rather, it's that polynomials in matrices (like $I+3A$) behave similarly to polynomials in real variables. But to invert a polynomial $P(x)$ for real $x$, you can just do a Taylor expansion on $\frac{1}{P(x)}$. Conceptually, I think this approach is simple and I like it because it's applicable to various situations. And it's rigorous too if at the end, you verify by direction multiplication, like I have done above. – Kim Jong Un Sep 14 '14 at 0:50
$A^2 + A + I$ is invertible (with inverse $I - A$), and $A^2$ is nilpotent (as $(A^2)^2 = A(A^3) = 0$), and the sum of a unit with a commuting nilpotent is again a unit.
Indeed, $((A^2 + A + I) + A^2)(I - A) = I + (I - A)A^2 = I + A^2$, which has inverse $I - A^2$, so $(2A^2 + A + I)^{-1} = (I - A)(I - A^2) = I - A - A^2$.
• Hm, I'm a bit confused. How did you come to the conclusion that $A^2 + A + I$ is invertible? – Alex Sep 14 '14 at 0:13
• @Alex: This comes from the identity (which holds in any ring with $1$) $1 - x^n = (1 - x)(x^{n-1} + x^{n-2} + \ldots + x + 1)$, which equals $1$ if $x^n = 0$ – zcn Sep 14 '14 at 2:07