Can't figure out necessary and sufficient conditions for showing that a cubic curve can be transformed to the quadratic one As i posted before Can a cubic Bezier curve be a quadratic one if two control points are equal from the cubic one?. I've got some great answers and they helped but i found another solution which involves matrix approach. Recalling the problem, which says if the transformation of cubic curve to quadratic (where two control points are equal) is possible. So i used degree reduction.
$$
\begin{bmatrix}
1 & 0 & 0 \\
1/3 & 2/3 & 0 \\
0 & 2/3 & 1/3 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
b_0 \\
b_1 \\
b_2
 \end{bmatrix}
=
\begin{bmatrix}
c_0 \\
c_1 \\
c_2 \\
c_3
\end{bmatrix}
$$
the equation $$DB=C$$ then multiplied by $D^T$
$$D^TDB=D^TC$$
then i get $B=(D^TD)^{-1}D^TC$. After that i found fundamental system of solutions of $D^T$
$$ \left[
\begin{array}{cccc|c}
  1 & 1/3 & 0 & 0 & 0\\
  0 & 2/3 & 2/3 & 0 & 0\\
  0 & 0 & 1/3 & 1 & 0
\end{array}
\right] $$
it gives the solution $(-1, 3, -3, 1)$ or $$-c_0+3c_1-3c_2+c_3=0$$
this condition tells if the points of the cubic curve satisfy this condition, then transformation of cubic to quadratic is possible. But as from my original condition of the problem (two control points of the cubic are equal) which means $c_1=c_2$ and thus it leads to $c_0=c_3$ (as condition $-c_0+3c_1-3c_2+c_3=0$ needs to be satisfied). It means cubic actually transforms to the linear curve (as we got the pairs $c_1=c_2$ and $c_0=c_3$). So as i figured $c_1=c_2$ and $c_0=c_3$ is actually sufficient condition for showing that cubic transforms to linear and it's impossible to transform it to quadratic with this condition. So i can't figure out how to find necessary condition and i'm also stuck for finding necessary and sufficient conditions for showing that transformation from cubic to quadratic via this $-c_0+3c_1-3c_2+c_3=0$ is possible. I will be grateful if somebody helps me with finding it
 A: My answer to your previous question already showed that the cubic is a quadratic if and only if 
$$
-c_0+3c_1-3c_2+c_3=0
$$
You have managed to prove this yourself, now (though it seems to me that your method is much more complicated than mine).
Now suppose that $c_1=c_2$, which is apparently what you mean by "two control points are equal". Substituting this in the equation above, we see that we have a quadratic if and only if
$$
-c_0+c_3=0
$$
In other words, a cubic with $c_1=c_2$ is a quadratic if and only if $c_0 = c_3$. In other words, the necessary and sufficient condition that you're seeking is $c_0 = c_3$.
The equation of the curve is then
$$
P(t) = (1-t)^3c_0 + 3t(1-t)^2c_1 + 3t^2(1-t)c_1 + t^3c_0
$$
A little algebra shows that this can be reduced to
$$
P(t0 = (1-t)^2 c_0 + 2t(1-t)\Bigl(\tfrac32 c_1 - \tfrac12 c_0 \Bigr) +t^2 c_0
$$
This is a quadratic Bezier curve with control points $c_0$, 
$\tfrac32 c_1 - \tfrac12 c_0$, $c_0$.
This curve happens to lie within the straight line between $c_0$ and $c_1$. In fact
as $t$ increases from $0$ to $1,$ the curve point $P(t)$ makes an out-and-back trip from $c_0$ to $\tfrac14(c_0+3c_1)$, and back to $c_0$ again.
So, some people might say that it's a "linear" curve. But, usually, a "linear" curve is defined to be one whose degree is equal to $1$. By that definition, our curve is not linear, it is quadratic, because its degree is equal to $2$.
