The non-existence of a matrix $A\in M_3(\mathbb{R})$ such that $A^2-A+I=0$.

Question

I'm trying to get my head around this problem, which states that no $3\times 3$ real-valued matrix can satisfy the equation $$A^2-A+I=0.$$ Of course, the problem is trivial if we consider characteristic polynomials and eigenvalues, but I want to prove it in the most elementary way possible, just basic theory of matrix operations and vector spaces. So considering the polynomial $p(λ):=\det (A-λI)$ without mentioning the word "eigenvalue" isn't good enough.

The only thing I can get out of this equation is that the matrix must be invertible and its inverse will be $A^{-1} = I-A$ and other deceptively useful identities, but I can't move on in any way.

Any hints are appreciated!

Answer 1

No such linear map exists. Suppose the contrary. Then $p(A)=0$ where $p(x)=x^2-x+1$. Pick any vector $v\ne0$. There are two possibilities:

1. $Av=cv$ for some $c\in\mathbb R$. Then $0=p(A)v=p(c)v$ and hence $p(c)=0$.
2. $V=\operatorname{span}\{v,Av\}$ is two-dimensional. Then $\mathbb R^3=V\oplus\operatorname{span}\{u\}$ for some $u\not\in V$ and $$ Au=v_1+cu $$ for some $v_1\in V$ and $c\in\mathbb R$. Since $A^2=A-I$, we have $A\{v,Av\}\subseteq V$. Hence $AV\subseteq V$ and in particular, $Av_1\in V$. It follows that \begin{align} A^2u=A(v_1+cu)=v_2+c^2u\tag{1} \end{align} for some $v_2\in V$. However, we also have \begin{align} A^2u=Au-u=v_1+(c-1)u.\tag{2} \end{align} By comparing the coefficients of $v$ in $(1)$ and $(2)$, we obtain $p(c)=0$.

Thus we arrive at contradictions in both cases, because $p(x)=0$ has not any real root. Hence $p(A)$ is never zero and $f$ does not exist.