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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 :)

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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.

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  • $\begingroup$ 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. $\endgroup$ – Alex Sep 14 '14 at 0:46
  • $\begingroup$ 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. $\endgroup$ – Kim Jong Un Sep 14 '14 at 0:50
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$A^2 + A + I$ 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 $(2A^2 + A + I)^{-1} = (I - A)(I - A^2) = I - A - A^2$.

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  • $\begingroup$ Hm, I'm a bit confused. How did you come to the conclusion that $A^2 + A + I$ is invertible? $\endgroup$ – Alex Sep 14 '14 at 0:13
  • $\begingroup$ @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 $x^n = 0$ $\endgroup$ – zcn Sep 14 '14 at 2:07

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