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Assuming a matrix $v^TA^T$ is compatible isn't the result of the transpose of their product $(vA)$?
In other words, if we assume $v^TA^T$ is compatible. What is the result of $(v^TA^T)^T$?
I did the following:
$(v^TA^T)^T=(v^T)^T(A^T)^T = vA$
But my answer is incorrect. What is the result? Thank you.
Your answer is indeed incorrect. You can see this by looking at the dimensions.
For example, let $v$ be a $n \times k$ matrix, and $A$ a $m \times n$ matrix, then $v^TA^T$ will be the product of a $(k \times n)$ matrix and a $(n \times m)$ matrix, resulting in a $k \times m$ matrix. No problem.
On the other hand, your answer would be the product of a $(n \times k)$ matrix and a $(m \times n)$ matrix, which is undefined unless $k = m$.
The correct answer is that $\left(v^TA^T\right)^T = Av$. You can work this out by looking at the value at an individual location in the product: the value at index $[i, j]$ is equal to: $$\begin{eqnarray}\left(v^TA^T\right)[i,j] = \sum_{l=1}^n v^T[i, l] A^T[l,j] &=& \sum_{l=1}^n v[l, i] A[j, l]\\&=& \sum_{l=1}^n A[j, l] v[l, i] \\ &=& (Av)[j,i] \\ &=& (Av)^T[i,j] \end{eqnarray}$$
When taking the transpose on a product, you replace the order. So: $$(AB)^t=B^tA^t$$ I believe this solves the mistake.
