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 🙂
Best Answer
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.