I was asked this question yesterday on my linear algebra qualifying exam (to the best of recollection):
Question 1: Find all $A\in \mathcal{M}_{n\times n}(\mathbb{R})$ such that for all $x\in \mathbb{R}^n$ there exists a $y\in \mathbb{R}^n$ satisfying \begin{equation} Ax=x+y. \end{equation}
The textbook we used was Golan's The Linear Algebra a Beginning Graduate Ought to Know, 3rd edition. In that book I found a similar exercise, p.145, #439, which states:
Question 2: Let $A = [a_{ij}] \in \mathcal{M}_{n\times n}(\mathbb{R})$ satisfy the condition that for each $v \in \mathbb{R}^n$ there exists a vector $y \in \mathbb{R}^n$ all entries in which are nonnegative satisfying $Av = v+y$. Show that $A = I$.
I did the question last and was rushed, but my work was something to the effect of:
Write $Ax=x+y$ as $$ (A-I)x=y. $$ Then,
1) Any matrix with eigenvalue $\lambda=1$, which has algebraic multiplicity equal to geometric multiplicity equal to $\dim(\mathbb{R}^n)=n$, and
2) A=I.
I retrospectively realize that (I think) these are equivalent.
With the relaxed hypothesis (i.e. no "nonnegative"), could you please help me with a solution to Question 1? Thank you in advance!
I believe that $A=I$ is the only matrix for which we can find such a y. As you have shown we know that $(A-I)x=y$. On the other hand we must have $A(2x)=2x+y$ which we can solve to $2(A-I)x=y$. This implies that $y=2y$ so $y$ must be the zero vector and $A$ must be the identity.