Preimage of non-invertible matrix I am given the matrix $$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}}  \\  \frac{-1}{\sqrt{2}}  &  \frac{1}{\sqrt{2}} \end{pmatrix}$$
Apparently this one is not invertible. Despite, is there a way to calculate the preimage of the convex hull of the points $\text{conv}((1,0),(-1,0),(0,2))$ (so the triangle spanned up by these three points) ?
If anything is unclear, please let me know.
 A: The kernel of the matrix is spanned by $\{(1,1)\}$, so we can add any real multiple of this vector to any point in the pre-image to get another point. This means if $(0,a)$ is in the pre-image, then so is every $(x,y)$ where $y=x+a$.
Computing the intersection of the range of the matrix, spanned by $\{(1,-1)\}$, with the three sides of the triangle, we get the points
$(0,0)$ and $\left(-\frac23,\frac23\right)$. A pre-image of these points is $(0,0)$ and $\left(0,\frac{2\sqrt2}3\right)$.
Thus, the pre-image would be the strip between
$$
y=x
$$
and
$$
y=x+\frac{2\sqrt2}3
$$
A: Write the matrix as $A = {1 \over \sqrt{2}}
\begin{bmatrix} 1 \\ -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \end{bmatrix}$, so we see that
${\cal R} A = \{ (x,-x) \}$.
Let $C$ be the convex set in question.
We have ${\cal R} A \cap C = \operatorname{co}\{(0,0)^T, (- {2 \over 3}, {2 \over 3})^T \}$. So, we need $(x,y)$ such that
$- {2 \over 3} \le {1 \over \sqrt{2}}(x-y) \le 0$.
That is, the points lying on and between the lines $x=y$ and $y=x+{2 \sqrt{2} \over 3}$.
Alternatively, note that $\ker A = \{(x,x) \}$ and ${\cal R} A \cap C = A ([-{2\sqrt{2} \over 3}, 0] \times \{0\})$, and so
$A^{-1}({\cal R} A \cap C) = [-{2\sqrt{2} \over 3}, 0] \times \{0\} + \ker A$.
A: Long story short, (I believe) the answer is as follows: Let $A$ denote the matrix in question.  Let $v_1,v_2,v_3$ refer to the points in question.  Let $y_1,y_2,y_3$ be any points satisfying $$
Ay_i = 
\begin{cases}
v_i & v_i \in \text{image}(A)\\
0   & \text{otherwise}
\end{cases}
$$
The preimage will consist of exactly the points that can be written in the form
$$
y = a + b
$$
where $a$ is in the convex hull of the points $y_i$ and $b$ is any point (column-vector) such that $Ab = 0$ (that is, $b \in \ker(A)$).
