If $a \neq b$, under what condition does $(I-C)a = (I-C)b$? Suppose $a, b \in \mathbb{R}^n$, and $a \neq b$ (that is, $a_j \neq b_j$ for at least one $j$, where $j = 1, \ldots, n)$.
Let $I$ denote a $n \times n$ identity matrix, and let $C \in \mathbb{R}^{n \times n}$.
I want to know whether it's possible for the following to hold:
$$(I-C)a = (I-C)b$$
Here's my attempt:
\begin{align*}
(I-C)a &= (I-C)b\\
\Leftrightarrow a - Ca &= b - Cb\\
\Leftrightarrow a - b &= Ca - Cb\\
\Leftrightarrow a-b &= C(a-b)
\end{align*}
Can I go further or is this it? So far, based on the above, I have $(I-C)a = (I-C)b$ iff $a-b = C(a-b)$.
 A: Your condition is equivalent to saying that one of the eigenvalues of $C$ is $1$, and that $a-b$ is in the eigenspace. If we start with a matrix $C$ that has $1$ as an eigenvalue, we can just pick any $v$ in this eigenspace and write $v=a-b$. For example, an easy matrix that has $1$ as an eigenvalue is:
$$
C = \begin{bmatrix}
1 & 0 \\ 0 & 2
\end{bmatrix}.
$$
A vector in the eigenspace of $1$ is $(1,0)$, so you could just take $a = (0,2)$ and $b=(1,2)$. You can then check that $(I-C)a = (I-C)b$.

Alternatively, you can start with your vectors $a,b$ and construct a matrix $C$. Since $a-b \neq 0$, we can add vectors to get a basis $\{a-b, v_2, ..., v_n\}$ for $\mathbb{R}^n$. We can complete to get another basis $\{a-b, w_2, ..., w_n\}$. Then define a linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ by setting $T(a-b) = a-b$ and $T(v_i) = w_i$ for $i \geq 2$. Let $C$ be the expression of $T$ in the standard basis, which is given by
$$
C = [a-b, w_2, ..., w_n] \cdot [a-b, v_2, ..., v_n]^{-1}.
$$
Then $C$ satisfies $C(a-b) = a-b$ since this is true of $T$; hence $(I-C)a = (I-C)b$. For example, take $a=(2,3), b=(2,1)$. Then $a-b = (0,2)$, and we can complete this to the bases $\{(0,2), (1,1)\}$ and $\{(0,2), (3,-1)\}$. Then:
$$
C =
\begin{bmatrix}
0 & 3 \\ 2 & -1
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\ 2 & 1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
3 & 0 \\ -2 & 1
\end{bmatrix}
$$
You can check that $C(a-b) = a-b$.
