I have read that the DCT can be computed as a matrix multiplication. The 8x8 DCT matrix is:
$D=\frac{1}{2}\left[\matrix{ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \cos \frac{\pi}{16} & \cos \frac{3\pi}{16} & \cos \frac{5\pi}{16} & \cos \frac{7\pi}{16} & \cos \frac{9\pi}{16} & \cos \frac{11\pi}{16} & \cos \frac{13\pi}{16} & \cos \frac{15\pi}{16} \\ \cos \frac{2\pi}{16} & \cos \frac{6\pi}{16} & \cos \frac{10\pi}{16} & \cos \frac{14\pi}{16} & \cos \frac{18\pi}{16} & \cos \frac{22\pi}{16} & \cos \frac{26\pi}{16} & \cos \frac{30\pi}{16} \\ \cos \frac{3\pi}{16} & \cos \frac{9\pi}{16} & \cos \frac{15\pi}{16} & \cos \frac{21\pi}{16} & \cos \frac{27\pi}{16} & \cos \frac{33\pi}{16} & \cos \frac{39\pi}{16} & \cos \frac{45\pi}{16} \\ \cos \frac{4\pi}{16} & \cos \frac{12\pi}{16} & \cos \frac{20\pi}{16} & \cos \frac{28\pi}{16} & \cos \frac{36\pi}{16} & \cos \frac{44\pi}{16} & \cos \frac{52\pi}{16} & \cos \frac{60\pi}{16} \\ \cos \frac{5\pi}{16} & \cos \frac{15\pi}{16} & \cos \frac{25\pi}{16} & \cos \frac{35\pi}{16} & \cos \frac{45\pi}{16} & \cos \frac{55\pi}{16} & \cos \frac{65\pi}{16} & \cos \frac{75\pi}{16} \\ \cos \frac{6\pi}{16} & \cos \frac{18\pi}{16} & \cos \frac{30\pi}{16} & \cos \frac{42\pi}{16} & \cos \frac{54\pi}{16} & \cos \frac{66\pi}{16} & \cos \frac{78\pi}{16} & \cos \frac{90\pi}{16} \\ \cos \frac{7\pi}{16} & \cos \frac{21\pi}{16} & \cos \frac{35\pi}{16} & \cos \frac{49\pi}{16} & \cos \frac{63\pi}{16} & \cos \frac{77\pi}{16} & \cos \frac{91\pi}{16} & \cos \frac{105\pi}{16} }\right]$
From linear algebra I thought that the DCT matrix was the matrix representation of the DCT as a linear operator. However, I think this is wrong, because the transform applied to a matrix $A$ is defined as:
$A'=DAD^T$
and the inverse transformation is:
$A=D^TA'D$
I know that $D$ is orthogonal, so $D^T=D^{-1}$, but I don't know why the transform is defined this way and two multiplications are needed instead of one. Obviously $D$ is not the matrix representation of the DCT, but then what is it? And why do we use its inverse to transform other matrices?
I am trying to understand as much as possible about the DCT, but from a linear algebra perspective. I haven't studied Fourier series yet, but I have at least the concepts.
It's because it's a 2-D transform. One works along the columns (the one on the left) and the other along the rows (the one on the right).