The definition of an eigenvalue:
$$ x * A = E * x $$ where $E$ is an eigenvalue, $A$ is a matrix and $x$ is a vector $≠ 0$.
My proof:
Let's assume that $E = -1$. Then $A * x = -x$. Adding $x$ to both sides and factoring:
$$x(A + Id) = 0$$
This part above feels illegal to me.
Continuing the proof, since $x ≠ 0$ and $(A + Id)$ has every entry positive, then $x(A + Id)$ can't be equal to $0$, so we've shown by contradiction that $E$ can't be equal to $-1$.
I'm mostly concerned about the part I marked as illegal. Can I just plug in $Id$, aka the identity matrix when factoring? If I can, can I do it every single time or are there limitations? If not, why not and is there any way to fix my proof easily?
Well but what about $A =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $x=[1,-1]^{\top}$? Doesn't the equation $Ax=-x$ hold? [or if you rather, $x=[1,-1]$ and $xA=-x$]
The error in your proof comes from your step saying that a square matrix $A'$ with all positive entries, namely, $A'=A+I$, is invertible. Equivalently, the error in your proof comes from your step saying that a square matrix $A'$ with all positive entries, is such that $yA'$ is nonzero for all nonzero $y$. That is clearly not true; consider $A' = A+I = \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}$. There is a vector $x$ such that the equation $xA'=0$ holds; in particular, take $x=[-1,1]$, then the equation $xA' = [-1,1]\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix} = [0,0]$ holds. [Equivalently, all entries of $A'$ are positive but $A'$ is not invertible. RECALL: A square matrix $A'$ is invertible iff $yA'$ is nonzero for all nonzero $y$.]