tonan6e

2022-09-07

How to multiply out brackets when they contain vectors

I am keen to know the general rule for any vector:

$(\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y}{)}^{T}$

How does it relate to $(x-y)(x-y)={x}^{2}-2xy+{y}^{2}$

Thanks for your help

I am keen to know the general rule for any vector:

$(\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y}{)}^{T}$

How does it relate to $(x-y)(x-y)={x}^{2}-2xy+{y}^{2}$

Thanks for your help

Abigayle Lynn

Beginner2022-09-08Added 12 answers

To see why removing brackets is a bit different, you can remove them step by step:

$(\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y}{)}^{T}=\mathbf{x}(\mathbf{x}-\mathbf{y}{)}^{T}-\mathbf{y}(\mathbf{x}-\mathbf{y}{)}^{T}=\mathbf{x}{\mathbf{x}}^{T}-\mathbf{x}{\mathbf{y}}^{T}-\mathbf{y}{\mathbf{x}}^{T}-\mathbf{y}{\mathbf{y}}^{T}$

The result cannot be simplified any further, because unlike the scalar case, $\mathbf{x}{\mathbf{y}}^{T}\ne \mathbf{y}{\mathbf{x}}^{T}$

$(\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y}{)}^{T}=\mathbf{x}(\mathbf{x}-\mathbf{y}{)}^{T}-\mathbf{y}(\mathbf{x}-\mathbf{y}{)}^{T}=\mathbf{x}{\mathbf{x}}^{T}-\mathbf{x}{\mathbf{y}}^{T}-\mathbf{y}{\mathbf{x}}^{T}-\mathbf{y}{\mathbf{y}}^{T}$

The result cannot be simplified any further, because unlike the scalar case, $\mathbf{x}{\mathbf{y}}^{T}\ne \mathbf{y}{\mathbf{x}}^{T}$

Deanna Gregory

Beginner2022-09-09Added 3 answers

EDIT/CAVEAT: I am interpreting the vectors as row vectors. Please see the other answer for interpreting the vectors as column vectors.

You're correct that you should treat it as the vector and its transpose. This is called an "inner product", or a "dot product". In the trivial case that the vectors are both scalars, it reduces to the formula you said. Otherwise, it becomes

$x{x}^{T}-x{y}^{T}-y{x}^{T}+y{y}^{T}$

$=|x{|}^{2}-2x\cdot y+|y{|}^{2}$

You're correct that you should treat it as the vector and its transpose. This is called an "inner product", or a "dot product". In the trivial case that the vectors are both scalars, it reduces to the formula you said. Otherwise, it becomes

$x{x}^{T}-x{y}^{T}-y{x}^{T}+y{y}^{T}$

$=|x{|}^{2}-2x\cdot y+|y{|}^{2}$

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