2

If matrix $A^3 = O$, then $A - I$ is nonsingular? True or false?

I tried to solve it like this:

Given a linear transformation $T: V \rightarrow V$, such that $$m(T) = A, m(O) = O, m(I) = I, m(T^3) = A^3$$ where $m(.)$ is an operator that converts a linear transformation into the matrix. Assume that $(T - I)(x) = O$. Further assume that $x \neq O$.

Then, $$(T - I)(x) = O \; \Rightarrow \; T(x) - I(x) = T(x) - x = O \; \Rightarrow \; T(x) = x$$ for some $x \neq O$. So, $T(x) = I(x)$ for some $x$.

But we also have $$m(T^3) = A^3 = O \; \Rightarrow \; T^3(x) = O \; \forall x \in V$$ so $T^3(x) = I^3(x) = O$ which is a contradiction. Thus, $x = O$.

Since $x = O$, the transformation $T - I$ is one-to-one ($(T - I)(x) = O$ implies $x = O$).

Thus, $m(T - I) = A - I$ is nonsingular.

Can someone check this proof? I found a solution online, which says this is an incorrect result.

I would also be interested to see other proofs but you are not allowed to used determinants (Apostol chapter 2 calculus vol.2 is the limiter)! You can freely use the isomorphism of the space of linear transformations and matrices with respect to addition, scalar multiplication, and composition.

John
  • 1,313

2 Answers2

1

Hint:

If $A^3=0$, what is $(A-I)(A^2+A+I)$?

J. W. Tanner
  • 60,406
1

Your proof is correct, but there is no need to work so hard. (Note also that there is no “operator that converts a linear operator to a matrix” in the abstract; you have to specify an ordered basis, and you get the coordinate matrix relative to that basis).

You can operate direction with the matrix $A$, multiplying $n\times 1$ vectors. You know that $A^3\mathbf{x}=\mathbf{0}$ for all $\mathbf{x}$. So if $(A-I)\mathbf{x}=\mathbf{0}$, then as you note you have $A\mathbf{x}=\mathbf{x}$, so $A^2\mathbf{x}=A\mathbf{x}=\mathbf{x}$, and $A^3\mathbf{x}=A\mathbf{x}=\mathbf{x}$. Since $A^3\mathbf{x}=\mathbf{0}$, that means $\mathbf{x}=\mathbf{0}$. This proves $A-I$ is nonsingular: we have shown that if $(A-I)\mathbf{x}=\mathbf{0}$, then $\mathbf{x}=\mathbf{0}$.

Arturo Magidin
  • 398,050