Syntax Explanation - Tranpose of Dot Product
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On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:
What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.
Last, how does w.x+b then become w hat . x hat?
linear-algebra vectors matrix-calculus
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
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add a comment |
$begingroup$
On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:
What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.
Last, how does w.x+b then become w hat . x hat?
linear-algebra vectors matrix-calculus
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
$endgroup$
add a comment |
$begingroup$
On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:
What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.
Last, how does w.x+b then become w hat . x hat?
linear-algebra vectors matrix-calculus
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
$endgroup$
On page 29 of https://arxiv.org/pdf/1802.01528.pdf there is syntax that I find confusing:
What does the syntax [wT, b]T mean? Is this a 2 dimensional matrix formed from a horizontal row w (transposed from column) and a column of b, all transposed? What is the final shape? Same question for x hat.
Last, how does w.x+b then become w hat . x hat?
linear-algebra vectors matrix-calculus
linear-algebra vectors matrix-calculus
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
asked Jan 11 at 21:54
Matthew ArthurMatthew Arthur
183
183
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
$$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
Similarly
$$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$
Finally,
$$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
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$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
add a comment |
$begingroup$
If we have
$$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
$$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
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2 Answers
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active
oldest
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2 Answers
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active
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$begingroup$
If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
$$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
Similarly
$$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$
Finally,
$$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
$endgroup$
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
add a comment |
$begingroup$
If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
$$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
Similarly
$$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$
Finally,
$$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
$endgroup$
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
add a comment |
$begingroup$
If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
$$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
Similarly
$$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$
Finally,
$$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
$endgroup$
If $mathbf{w} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_nend{bmatrix}$ then
$$hat{mathbf{w}} = begin{bmatrix}w_1 \ w_2 \ vdots \ w_n \ b end{bmatrix}.$$
Similarly
$$hat{mathbf{x}} = begin{bmatrix}x_1 \ vdots \ x_n \ 1end{bmatrix}.$$
Finally,
$$hat{mathbf{w}} cdot hat{mathbf{x}} = w_1 x_1 + w_2 x_2 + cdots + w_n x_n + b =mathbf{w} cdot mathbf{x} + b.$$
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
answered Jan 11 at 22:01
angryavianangryavian
40.5k23280
40.5k23280
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
add a comment |
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
$begingroup$
How does the Transpose play into this?
$endgroup$
– Matthew Arthur
Jan 11 at 22:02
1
1
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
$begingroup$
@MatthewArthur Typically a vector is assumed to be a column vector. They transpose $mathbf{w}$ to make it a row vector in order to use the "horizontal" notation $[mathbf{w}, b] = [w_1, ldots, w_n, b]$, and then transpose it again to make the whole thing a column vector again.
$endgroup$
– angryavian
Jan 11 at 22:05
add a comment |
$begingroup$
If we have
$$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
$$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
$endgroup$
add a comment |
$begingroup$
If we have
$$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
$$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
$endgroup$
add a comment |
$begingroup$
If we have
$$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
$$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
$endgroup$
If we have
$$w = begin{bmatrix} w_{1}\w_{2}\vdots\w_{n}end{bmatrix}$$
then $w^{T} = [w_{1}, ldots, w_{n}]$, $[w^{T},b] = [w_{1}, ldots, w_n, b]$, and
$$[w^{T},b] = [w_{1}, ldots, w_n, b]^{T} =begin{bmatrix}w_{1}\vdots\w_{n}\b end{bmatrix}.$$
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
answered Jan 11 at 22:06
Morgan RodgersMorgan Rodgers
9,66721439
9,66721439
add a comment |
add a comment |
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