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From this question, does $\textbf{a}\cdot\textbf{b}=\textbf{a}^T\textbf{b}$?

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  • $\begingroup$ It occurs to me that despite my answer, there is a subtle difference between them: $\mathbf a \cdot \mathbf b$ is a scalar, that is, an element of the underlying structure of the matrix space, but $\mathbf a^T \mathbf b$ is an order $1$ matrix. The two behave similarly, but technically speaking you can't multiply an $n \times m$ matrix by a $1 \times 1$ matrix using conventional matrix multiplication -- but you can multiply an $n \times m$ matrix by a scalar. $\endgroup$ – Prime Mover Jun 26 at 9:17
  • $\begingroup$ @PrimeMover As I have observed many times, there is a distinction between a scalar $x$ and the $1\times 1$ matrix $(x)$, but maintaining that distinction is in practice virtually impossible :-) $\endgroup$ – Angina Seng Jun 26 at 9:21
  • $\begingroup$ @AnginaSeng Maintaining the distinction is probably not important. Being aware of it is another thing altogether, and it probably is important at least to understand the distinction. $\endgroup$ – Prime Mover Jun 26 at 10:39
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Assuming $\mathbf a$ and $\mathbf b$ are column vectors, then yes.

The transpose of a column vector is a row vector.

Then the identity trivially follows from the definitions of transpose and matrix product.

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  • $\begingroup$ Thank you so much! $\endgroup$ – J. Doe Jun 26 at 9:11

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