91Ó°ÊÓ

Let A be an m x n matrix and let B be the resulting matrix after multiplying A's transpose by A, which will be an n x n matrix. Recall that for two matrices C and D, the entry in the i-th row and j-th column of the product CD can be found as follows: (CD)_{ij} = Sum of products of the corresponding entries in the i-th row of C and the j-th column of D Now, let's apply this rule to our problem, where C is A's transpose and D is A: \(b_{ij}\) = Sum of products of the corresponding entries in the i-th row of \(A^T\) and the j-th column of A Since A's transpose has the same columns as the rows of A, the i-th row of \(A^T\) corresponds to the i-th row of A, and the j-th column of A corresponds to the j-th row of A. So, we can rewrite the equation as: \(b_{ij}\) = Sum of products of the corresponding entries in the i-th row of A and the j-th row of A

Now, let's consider the dot product of two vectors. Given two vectors \(v_i\) and \(v_j\), their dot product can be calculated as follows: \(v_i \cdot v_j\) = Sum of products of the corresponding entries in \(v_i\) and \(v_j\) Notice that this operation is quite similar to the one we performed in the matrix multiplication step. Taking the corresponding entries of two vectors and summing their products is exactly the same as taking the corresponding entries in the i-th row of A and the j-th row of A and summing their products. Therefore, we can write: \(b_{ij}\) = \(\mathbf{a}_i^T \mathbf{a}_j\) Where \(\mathbf{a}_i\) and \(\mathbf{a}_j\) represent the i-th and j-th row vectors of A, respectively. So, this shows that the entries of matrix B, which results from multiplying the transpose of matrix A with matrix A, can be represented as the dot products of the row vectors of matrix A.