# Does $\textbf{a}\cdot\textbf{b}=\textbf{a}^T\textbf{b}$?

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

• 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. – Prime Mover Jun 26 at 9:17
• @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 :-) – Angina Seng Jun 26 at 9:21
• @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. – Prime Mover Jun 26 at 10:39

Assuming $$\mathbf a$$ and $$\mathbf b$$ are column vectors, then yes.