I was going through the camera matrix explained in the wikipedia article and understand how the matrix K \begin{bmatrix}f_x&s&x_0\\0&f_y&y_0\\0&0&1\end{bmatrix} is built. The projection matrix is then essentially K * [R | T]

However, I am not able to understand what the perspective projection matrix is and how is it same to the intrinsic matrix K

  • $\begingroup$ "The **projection matrix is then essentially...". AFAICS That's not a projection matrix.. it's not really a "2D" rotation + translation matrix either. Are you familiar with homogeneous coordinates? You really need to understand those first. $\endgroup$
    – Simon F
    Commented Jan 14, 2020 at 8:40
  • 1
    $\begingroup$ @SimonF this matrix can be easily broken into 2D Scaling "+" 2D translation. Also, this is essentially the simple perspective projection matrix. (multiply by it and then homogenise the co-ordinates.) $\endgroup$
    – midi
    Commented Jan 14, 2020 at 17:28
  • 1
    $\begingroup$ @SimonF I now understand that the projection matrix (with the viewing frustum mapped to a cube) is used as it contains depth information which is lost in the simple perspective projection matrix I had written earlier. $\endgroup$
    – midi
    Commented Jan 14, 2020 at 17:30
  • $\begingroup$ Oh good. :-) FWIW my first comment was that you'd written [R|T] which seemed to me to imply Rotation +Scaling, but the matrix requires another term. $\endgroup$
    – Simon F
    Commented Jan 15, 2020 at 9:42

1 Answer 1


The projection matrix is then essentially K * [R | T]

Is incorrect, actually the K Matrix is similar to projection matrix and the R|T is called camera transform matrix or view transform matrix Or called extrinsic matrix. So focus on the projection stuff and forget about camera transform, what is the difference between K Matrix and Perspective Projection Matrix(call it P Matrix later)?

  • For K Matrix it transform 3D points to 2D pixels in image space.
  • For P Matrix it transform 3D points to NDC space.

Take a look at two matrices:

$$K = \begin{bmatrix}f_x& 0& c_x\\ 0& f_y& c_y\\ 0 & 0 & 1\\\end{bmatrix}$$

$$P = \begin{bmatrix} \frac{1}{t*a}& 0& 0& 0\\0& \frac{1}{t}& 0& 0\\ 0 & 0 & A& B&\\ 0 & 0 & -1& 0\\ \end{bmatrix}$$ $$t=tan(\frac{fovy}{2})$$ $$a=\frac{width}{height}$$

Let's add perspective divide and show the result of the above two matrices:

Intrinsic case: $$x_{2d} = \frac{x_0}{z_0 * \frac{1}{f_x}} + c_x$$

Perspective case: $$x_{2d} = \frac{x_0}{-z_0(t*a)}$$

Similar with some difference.

The image space: Origin from left-top corner so should add Cx Cy as the offset from center to left-top corner. And in NDC space we assume Z-axis direct out of screen so P(3,2) = -1.


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