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I am reading this answer https://math.stackexchange.com/a/186254/130408. The original question in that post is about deducing the projection matrix. But I have difficulties in understanding that answer.

The answerer uses 4 points to construct a frame first, $$ A= \left[\begin{array}{ccc}x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\\x_4&y_4&z_4&1\end{array}\right] $$

and then finds the coordinate of the 5th point under this frame, $$ \left[\begin{array}{ccc}d_1&d_2&d_3&d_4\end{array}\right]=\left[\begin{array}{ccc}x_5&y_5&z_5&1\end{array}\right]\left[\begin{array}{ccc}x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\\x_4&y_4&z_4&1\end{array}\right]^{-1} $$

Finally this "local coordinate" of the 5th point is used as a weight to scale the original frame to get the projective frame, $$ M_{[x\;y\;z]}=\left[\begin{array}{ccc}d_1x_1&d_1y_1&d_1z_1&d_1\\d_2x_2&d_2y_2&d_2z_2&d_2\\d_3x_3&d_3y_3&d_3z_3&d_3\\d_4x_4&d_4y_4&d_4z_4&d_4\end{array}\right] $$

I know that a 4D projective frame has 15 DOFs, therefore 5 pairs of points(totally 3 * 5 = 15 equations) are necessary.

But why the first frame $A$ is not a valid projective frame? I prefer an answer in terms of the extra requirements of $A$ to be a valid projective frame, which I have to take care of when I embed 3D points in projective space.

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In this answer I've outlined the same approach, how to obtain a projection matrix, only I did so for the projective plane not three-space. The approach is very much the same, though. Perhaps my slightly different notation and my somewhat more verbose explanation will help you, particularly if taken together with robjohn's answer.

In essence, the key issue here is uniqueness. In homogeneous coordinates, a multiple of a given coordinate vector represents the same point. So the representant isn't unique, you are free to choose any scalar multiple. Likewise for the matrix as a whole.

Plugging four points into the matrix (as rows in robjohn's notation, as columns in mine), you get a matrix which will map the unit vectors of $\mathbb R^4$ to those four points. But if you scale the points differently, the matrix would still map the unit vectors to these four points (only other representants of them), but on the whole represent a different transformation. So four points can be mapped to the desired locations (so in a certain sense it is quite “valid”), but the mapping isn't unique. To obtain uniqueness, you introduce a fifth point, namely the image of $[1,1,1,1]$. Now your matrix is unique up to a scalar factor which has no geometric significance. That's the reason why using five points is the right thing to do in projective three-space.

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