7
$\begingroup$

I came across the symbol $\otimes$ as below and I would like to know what this symbol $\otimes$ means:

$\text{.... the projection operator P is given by: }$ $$P = I_nd - \nabla G^T(\nabla G \nabla G^T)^{-1} \nabla G= I_{nd} - I_d \otimes uu^T,$$ where $I_x$ denotes the identity matrix of size $x\times x$ and $\mathbf{u}$ is the unit vector , $\mathbf{u} = (1,1,1,\dots,1)/\sqrt{n})$ in $\mathbb{R}^n$

$\endgroup$
9
$\begingroup$

In this context, $\otimes$ refers specifically to the Kronecker product. In particular, we have $$ I_d \otimes B = \overbrace{B \oplus B \oplus \dots \oplus B}^d = \text{diag}(\overbrace{B,B, \dots, B}^d)\\ = \pmatrix{B\\&B\\&&\ddots\\&&&B} $$

$\endgroup$
4
$\begingroup$

Here is an official reference, more or less, from Dummit and Foote's Abstract Algebra (3rd. ed., p. 421):

Let $A=(\alpha_{ij})$ and $B$ be $r\times n$ and $s\times m$ matrices, respectively, with coefficients from any commutative ring. The Kronecker product or tensor product of $A$ and $B$, denoted by $A\otimes B$, is the $rs\times nm$ matrix consisting of an $r\times n$ block matrix whose $i, j$ block is the $s\times m$ matrix $\alpha_{ij} B$.

$\endgroup$
0
$\begingroup$

It means the tensor product...

$\endgroup$
0
$\begingroup$

Tensor Product. Check here for reference. http://en.wikipedia.org/wiki/Tensor_product

EDIT: Ninja'd

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.