Prove that $\det(I_n+AB)\neq0$

Question

Let's consider $A, B\in M_n(\mathbb C)$ such that $$A+B=I_n$$ and $$A^2=A^3$$ Prove that $$\det(I_n+AB)\neq0$$

From $A+B=I_n$ we have that $A^3+A^2B=A^2$, so $A^2B=0_n$

I supposed that $\det(I_n+AB)=0$, i.e. $$\det(I_n+A-A^2)=0$$ $$\det((I_n+A-A^2)A)=0$$ $$\det(A)=0$$ $$\det((I_n+A-A^2)(A-I_n))=0$$ $$\det(A^2-I_n)=0$$ $$\det((A^2-I_n)(A-I_n))=0$$ $$\det(A-I_n)=0$$

So far we have $\det(A)=\det(A-I_n)=\det(I_n+A-A^2)=0$ and $A^2(I_n-A)=0_n$ and I don't know how to find the mistake (I supposed that $\det(I_n+AB)=0$ ).

Note that $A$ and $B$ commute. As you have shown, $A^2B=0$. Hence $(I+AB)(I-AB)=I-A^2B^2=I$ and we are done. — Kelenner, Feb 10 '18 at 08:51

score 2 · Answer 1 · answered Feb 10 '18 at 08:43

2

From $A^2=A^3$ the only possible eigenvalues of $A$ are $0$ and $1$.

As you say $I+AB=I+A-A^2$, but the only possible eigenvalues of $I+A-A^2$ are $1+0-0^2=1$ and $1+1-1^2=1$. So $I+AB$ is invertible.

answered Feb 10 '18 at 08:43

Angina Seng

158,341

How do we know that all the eigenvalues of $I+A-A^2$ come from eigenvalues of $A$, i.e. are of the form $1+\lambda -\lambda^2$ where $\lambda$ is an eigenvalue of $A$ ? – Ewan Delanoy Feb 10 '18 at 08:46
@EwanDelanoy think about Jordan form. – Angina Seng Feb 10 '18 at 08:56
I see now : if $A$ has a Jordan form in some basis $B$, then $I+A-A^2$ also has a Jordan form in $B$. Thanks for the explanation – Ewan Delanoy Feb 10 '18 at 09:03

score 1 · Accepted Answer · answered Feb 10 '18 at 08:52

Suppose, by contradiction, that $I+AB$ is singular. Then, there is a nonzero vector $e$ such that $(I+AB)(e)=0$, which means that $(1) A^2e=e+Ae$. Applying $A$, we deduce $A^3e=Ae+A^2e=e+2Ae$. But $A^3=A^2$, so $e+Ae=e+2Ae$ and we deduce $Ae=0$. But then $A^2e=0$, and from (1) we deduce $Ae=-e$, so $e=0$ which is impossible.

Prove that $\det(I_n+AB)\neq0$

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