# Why does this happen: $(v^TA^T)^T\ne vA$

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: $$\begin{eqnarray}\left(v^TA^T\right)[i,j] = \sum_{l=1}^n v^T[i, l] A^T[l,j] &=& \sum_{l=1}^n v[l, i] A[j, l]\\&=& \sum_{l=1}^n A[j, l] v[l, i] \\ &=& (Av)[j,i] \\ &=& (Av)^T[i,j] \end{eqnarray}$$

• Thank you. Working it out like this helped me better understand. Commented Oct 3, 2020 at 5:56

$$(v^TA^T)^T=(A^T)^T(v^T)^T=Av$$.

When taking the transpose on a product, you replace the order. So: $$(AB)^t=B^tA^t$$ I believe this solves the mistake.