I've read in multiple places that DCT decorrelates Toeplitz matrices and images usually have Toeplitz structure. Can you explain with an example how DCT decorrelates a Toeplitz matrix?
Example for DFT:
DFT
DFT decorrelates circular matrices. This is how I was able to understand that.
Suppose $X$ is a matrix whose correlation matrix is not diagonal. We want to find a transformation $Y=AX$ such that correlation matrix of $Y$ is diagonal.
$\mathbb{E}[YY^T] = \mathbb{E}[AXX^TA^T] = A\mathbb{E}[XX^T]A^T$
$$\mathbb{E}[YY^T] = \mathbb{E}[AXX^TA^T] = A\mathbb{E}[XX^T]A^T$$
Let the Eigen Value Decomposition be $\mathbb{E}[XX^T]=U\Lambda U^T$
$$\mathbb{E}[XX^T]=U\Lambda U^T$$
Then $A=U^T$ gives, $\mathbb{E}[YY^T]=U^TU\Lambda U^TU = \Lambda$,
$$A=U^T \qquad\text{gives}\qquad\mathbb{E}[YY^T]=U^TU\Lambda U^TU = \Lambda$$ which is diagonal.
So given a matrix, its eigenvector matrix decorrelates it.
Consider a circular matrix $$A = \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \\ \end{bmatrix}$$ A $3 \times 3$ DFT matrix is given by $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w^2 & w \\ \end{bmatrix}$$ and $w^3 = 1$.$$\begin{bmatrix} 1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w^2 & w \\ \end{bmatrix}\qquad\text{and}\qquad w^3 = 1 $$
We can easily see that all the columns (or rows since it is symmetric) of the above matrix are eigenvectors of the considered circular matrix $A$. Thus DFT decorrelates circular matrices.
Is it possible to show in a similar way that DCT decorrelates a Toeplitz matrix?
