Outer product of row vectors. Does $\mathbf{x}^T \otimes \mathbf{y}^T$ = $\mathbf{x} \otimes \mathbf{y}$?

Question

Does $\mathbf{x}^T \otimes \mathbf{y}^T$ = $\mathbf{x} \otimes \mathbf{y}$?

Answer 1

The outer product is equivalent to the standard matrix product if we interpret (column) vectors as $n \times 1$ matrices: $$ \mathbf{x} \otimes \mathbf{y} = \mathbf{x} \, \mathbf{y}^\top. $$ You can see by looking at dimensions that this makes sense. In order for matrix multiplication to be defined the column count on the left must match the row count on the right, and here we have $$ (n \times 1)(1 \times n) = (n \times n). $$

On the other hand, $\mathbf{x}^\top \otimes \mathbf{y}^\top$ doesn't make sense as an outer product, as $\mathbf{x}^\top$ and $\mathbf{y}^\top$ are row vectors (often interpreted as covectors in many contexts). If we just go ahead and blindly compute $$ \mathbf{x}^\top \otimes \mathbf{y}^\top = \mathbf{x}^\top (\mathbf{y}^\top)^\top = \mathbf{x}^\top \mathbf{y}, $$ what we have is a row vector multiplied by a column vector, which yields a scalar since $$ (1 \times n)(n \times 1) = (1 \times 1). $$ This is called the inner product of $\mathbf{x}$ and $\mathbf{y}$.

Answer 2

Without making matrix identifications, if $x,y\in V$, then $x^\top,y^\top\in V^*$. Thus, $x\otimes y\in V\otimes V$ and $x^\top\otimes y^\top\in V^*\otimes V^*$. Although both of these can be identified with $V\otimes V^*\cong \text{Hom}(V,V)$ once you choose a basis and dual basis, these are very different entities.