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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.

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3 Answers 3

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

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  • $\begingroup$ Thank you. Working it out like this helped me better understand. $\endgroup$
    – martinbshp
    Oct 3, 2020 at 5:56
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$(v^TA^T)^T=(A^T)^T(v^T)^T=Av$.

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

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