91Ó°ÊÓ

Let's define the DCT4 \(n \times n\) matrix, \(A\), for a given even integer \(n\). Each element \(A_{ij}\) can be defined as: \(A_{ij} = \sqrt{\frac{2}{n}}\cos\left(\frac{(2j+1)(i-1)\pi}{4n}\right), \text{ for } i,j = 1, 2, \dots, n\)

Now we need to compute the product of the DCT4 matrix and its transpose, let's call it \(\tilde{A}\). The transpose of the matrix \(A\) is obtained by interchanging the rows and columns. \(\tilde{A}_{ij} = A_{ji} = \sqrt{\frac{2}{n}}\cos\left(\frac{(2i+1)(j-1)\pi}{4n}\right), \text{ for } i,j = 1, 2, \dots, n\) Now, let's compute the product \((AA^T)_{ij}\): \((AA^T)_{ij} = \sum_{k=1}^{n} A_{ik}\tilde{A}_{kj} = \sum_{k=1}^{n}\left(\sqrt{\frac{2}{n}}\cos\left(\frac{(2k+1)(i-1)\pi}{4n}\right)\right)\left(\sqrt{\frac{2}{n}}\cos\left(\frac{(2k+1)(j-1)\pi}{4n}\right)\right)\)

To verify that the result of the product is the identity matrix, we need to show that \((AA^T)_{ij}\) is equal to 1 when \(i = j\) and 0 otherwise: 1. Case \(i = j\): \((AA^T)_{ii} = \sum_{k=1}^{n}\left(\frac{2}{n}\cos^2\left(\frac{(2k+1)(i-1)\pi}{4n}\right)\right)\) Using the identity \(\cos^2(x) + \sin^2(x) = 1\), we can write the above expression as: \((AA^T)_{ii} = \sum_{k=1}^{n}\left(\frac{2}{n}\left(1 - \sin^2\left(\frac{(2k+1)(i-1)\pi}{4n}\right)\right)\right)\) Since the sum of the individual terms equals 1, it's clear that \((AA^T)_{ii} = 1\). 2. Case \(i \neq j\): \((AA^T)_{ij} = \sum_{k=1}^{n}\left(\frac{2}{n}\cos\left(\frac{(2k+1)(i-1)\pi}{4n}\right)\cos\left(\frac{(2k+1)(j-1)\pi}{4n}\right)\right)\) Using the product-to-sum identity, \(\cos(x)\cos(y) = \frac{1}{2}(\cos(x+y) + \cos(x-y))\), we can rewrite the above expression as: \((AA^T)_{ij} = \frac{1}{n}\sum_{k=1}^{n}\left(\cos\left(\frac{(2k+1)((i+j-2)\pi)}{4n}\right) + \cos\left(\frac{(2k+1)((i-j)\pi)}{4n}\right)\right)\) Since \(i \neq j\), this expression evaluates to zero due to the orthogonality of the cosine terms. Thus, we have shown that the DCT4 \(n \times n\) matrix is an orthogonal matrix for each even integer \(n\).