How do you prove (or disprove) the statement: If $A^3 = 0$, then A-I is non-singular

Question

I've proven something similar: A*A =0, then A + I is non-singular for 2x2 matrices. But not sure how to proceed for $A^3 = 0$, then A-I is non-singular

Also, not sure how to prove A*A =0, then A + I is non-singular for $nxn$ matrices.

(Note: all matrices are $nxn$ matrices, and I is the identity matrix).

Answer 1

$A^3=0$ will mean that the annihilating polynomial of $A$ is $x^3=0$ and minimal polynomial divides annihilating polynomial hence its only roots are $0$.

Thus $0$ is the only eigen value of $A$.

Claim: $-1$ is the only eigen value of $A-I$(Why?)

Let $\lambda $ be an eigen value of $A-I$ and $v$ the corresponding eigen vector

$(A-I)v=\lambda v\implies Av-v=\lambda v\implies Av=(\lambda+1)v$

Thus $\lambda+1$ is an eigen value of $A$ but $0$ is its only eigen value thus $\lambda+1=0\implies \lambda=-1$

Thus $\det(A-I)=$product of eigen values $=(-1)^n$ where $n=$ order of $A$

Thus $A-I$ is non-singular

Answer 2

The are already some nice proofs in the other answer and the comments. Here is an alternative proof by reductio ad absurdum.

A square matrix $M$ is nonsingular if and only if the equation $Mx=0$ has only the trivial solution. So, suppose $(A-I)x=0$, i.e. $x=Ax$. It suffices to prove that $x=0$.

Since $A^3=0$, from $x=Ax$ we get $A^2x=A^3x=0$.

Since $A^2x=0$, from $x=Ax$ we get $Ax=A^2x=0$.

Since $Ax=0$, from $x=Ax$ we get $x=0$. QED

Answer 3

Per binomial identities $$ (I+A+A^2)·(I-A)=I^3-A^3=I $$ so that the inverse matrix to $(I-A)$ is easily identifiable.