1. Let's state the problem: We want to understand why $$y^T A = (A^T y)^T$$ holds, where $y$ is a vector and $A$ is a matrix. 2. Recall that for a vector $y$ (column vector), $y^T$ is a row vector. 3. Consider the product $y^T A$: Since $y^T$ is $1 \times n$ and $A$ is $n \times m$, the result is a $1 \times m$ row vector. 4. Now look at the right-hand side: $(A^T y)^T$. - $A^T$ is an $m \times n$ matrix. - Multiplying $A^T y$ gives an $m \times 1$ column vector. - Taking the transpose $\bigl((A^T y)^T\bigr)$ yields a $1 \times m$ row vector. 5. The key is to compare the components. For the $j$-th entry, $$ (y^T A)_j = \sum_{i} y_i A_{ij} $$ $$ \bigl((A^T y)^T\bigr)_j = (A^T y)_j = \sum_{i} A^T_{ji} y_i = \sum_{i} A_{ij} y_i $$ 6. Since both sums are identical, the row vectors are equal. 7. Thus, we have shown that $$ y^T A = (A^T y)^T $$ This equality is due to the way matrix multiplication and transposition work, preserving the vector and matrix component order appropriately.