How does tensor product/multiplication work?

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In Tensorflow, I saw the following example:

mat_a = tf.constant(np.arange(1,12, dtype=np.int32), shape=[2,2,3]) 
mat_b = tf.constant(np.arange(12,24, dtype=np.int32), shape=[2,3,2]) 
mul_c = tf.matmul(mat_a, mat_b)

with tf.Session() as sess: 
 runop = sess.run(mul_c) 
print(runop) 
[[[ 88 94] 
 [214 229]] 
 [[484 508] 
 [642 674]]]

How does the tensor multiplication work?

edited 48 mins ago

Stephen Rauch

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asked 2 hours ago

Hendo Ley

111

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up vote
2
down vote

favorite

In Tensorflow, I saw the following example:

mat_a = tf.constant(np.arange(1,12, dtype=np.int32), shape=[2,2,3]) 
mat_b = tf.constant(np.arange(12,24, dtype=np.int32), shape=[2,3,2]) 
mul_c = tf.matmul(mat_a, mat_b)

with tf.Session() as sess: 
 runop = sess.run(mul_c) 
print(runop) 
[[[ 88 94] 
 [214 229]] 
 [[484 508] 
 [642 674]]]

How does the tensor multiplication work?

edited 48 mins ago

Stephen Rauch

1,29541128

asked 2 hours ago

Hendo Ley

111

New contributor

add a commentÂ |Â

up vote
2
down vote

favorite

In Tensorflow, I saw the following example:

mat_a = tf.constant(np.arange(1,12, dtype=np.int32), shape=[2,2,3]) 
mat_b = tf.constant(np.arange(12,24, dtype=np.int32), shape=[2,3,2]) 
mul_c = tf.matmul(mat_a, mat_b)

with tf.Session() as sess: 
 runop = sess.run(mul_c) 
print(runop) 
[[[ 88 94] 
 [214 229]] 
 [[484 508] 
 [642 674]]]

How does the tensor multiplication work?

edited 48 mins ago

Stephen Rauch

1,29541128

asked 2 hours ago

Hendo Ley

111

New contributor

In Tensorflow, I saw the following example:

mat_a = tf.constant(np.arange(1,12, dtype=np.int32), shape=[2,2,3]) 
mat_b = tf.constant(np.arange(12,24, dtype=np.int32), shape=[2,3,2]) 
mul_c = tf.matmul(mat_a, mat_b)

with tf.Session() as sess: 
 runop = sess.run(mul_c) 
print(runop) 
[[[ 88 94] 
 [214 229]] 
 [[484 508] 
 [642 674]]]

How does the tensor multiplication work?

tensorflow linear-algebra

edited 48 mins ago

Stephen Rauch

1,29541128

asked 2 hours ago

Hendo Ley

111

New contributor

edited 48 mins ago

Stephen Rauch

1,29541128

asked 2 hours ago

Hendo Ley

111

New contributor

edited 48 mins ago

Stephen Rauch

1,29541128

edited 48 mins ago

Stephen Rauch

1,29541128

edited 48 mins ago

Stephen Rauch

1,29541128

asked 2 hours ago

Hendo Ley

111

New contributor

asked 2 hours ago

Hendo Ley

111

asked 2 hours ago

Hendo Ley

111

New contributor

Hendo Ley is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

You may want to read the documentation.

output[..., i, j] = sum_k (a[..., i, k] * b[..., k, j]), for all indices i, j.

For instance, in your example

$~~88=1times12+2times14+3times16,~~~94=1times13+2times15+3times17$
$214=4times12+5times14+6times16,~229=4times13+5times15+6times17$

answered 1 hour ago

user12075

73427

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up vote
0
down vote

Tensor multiplication is just a generalization of matrix multiplication which is just a generalization of vector multiplication.

Matrix multiplication is defined as:

$$ A_i cdot B_j = C_i, j$$

where $i$ is the $i^th$ row, $j$ is the $j^th$ column, and $cdot$ is the dot product. Therefore it just a series of dot products.

One can then see how this extends to tensors:
$$mathbfA_i cdot mathbfB_j = mathbfC_i, j $$

where $i$ is the $i^th$ row-wise matrix of the tensor, and $j$ is the $j^th$ column-wise matrix of the tensor... and is therefore just a series of matrix multiplications - or a series of a series of dot products.

Assuming all tensors are of rank three(it can be described with three coordinates):

$$mathbfA otimes mathbfB = mathbfA_i, j cdot mathbfB_j, k = mathbfC_i, j, k$$

which means the $(i,j)^th$ vector of $mathbfA$ times the $(j, k)^th$ vector of $mathbfB$.

answered 17 mins ago

Daniel

1137

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up vote
0
down vote

I'll give you a small example, if you do the following Kronecker product
beginequation
beginbmatrix
colorred1 \
colorgreen5 \
colorblue10
endbmatrix
otimes
beginbmatrix
2 \
4
endbmatrix
=
beginbmatrix
colorred1 beginbmatrix
2 \
4
endbmatrix \\
colorgreen5 beginbmatrix
2 \
4
endbmatrix \\
colorblue10 beginbmatrix
2 \
4
endbmatrix \
endbmatrix
=
beginbmatrix
2 \
4 \
10 \
20 \
20 \
40
endbmatrix
endequation
The Kronecker product works the same way for matrices as well.

answered 7 mins ago

Ahmad Bazzi

1011

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

You may want to read the documentation.

output[..., i, j] = sum_k (a[..., i, k] * b[..., k, j]), for all indices i, j.

For instance, in your example

$~~88=1times12+2times14+3times16,~~~94=1times13+2times15+3times17$
$214=4times12+5times14+6times16,~229=4times13+5times15+6times17$

answered 1 hour ago

user12075

73427

add a commentÂ |Â

up vote
1
down vote

You may want to read the documentation.

output[..., i, j] = sum_k (a[..., i, k] * b[..., k, j]), for all indices i, j.

For instance, in your example

$~~88=1times12+2times14+3times16,~~~94=1times13+2times15+3times17$
$214=4times12+5times14+6times16,~229=4times13+5times15+6times17$

answered 1 hour ago

user12075

73427

add a commentÂ |Â

up vote
1
down vote

You may want to read the documentation.

output[..., i, j] = sum_k (a[..., i, k] * b[..., k, j]), for all indices i, j.

For instance, in your example

$~~88=1times12+2times14+3times16,~~~94=1times13+2times15+3times17$
$214=4times12+5times14+6times16,~229=4times13+5times15+6times17$

answered 1 hour ago

user12075

73427

You may want to read the documentation.

output[..., i, j] = sum_k (a[..., i, k] * b[..., k, j]), for all indices i, j.

For instance, in your example

$~~88=1times12+2times14+3times16,~~~94=1times13+2times15+3times17$
$214=4times12+5times14+6times16,~229=4times13+5times15+6times17$

answered 1 hour ago

user12075

73427

answered 1 hour ago

user12075

73427

answered 1 hour ago

user12075

73427

answered 1 hour ago

user12075

73427

add a commentÂ |Â

up vote
0
down vote

Tensor multiplication is just a generalization of matrix multiplication which is just a generalization of vector multiplication.

Matrix multiplication is defined as:

$$ A_i cdot B_j = C_i, j$$

where $i$ is the $i^th$ row, $j$ is the $j^th$ column, and $cdot$ is the dot product. Therefore it just a series of dot products.

One can then see how this extends to tensors:
$$mathbfA_i cdot mathbfB_j = mathbfC_i, j $$

Assuming all tensors are of rank three(it can be described with three coordinates):

$$mathbfA otimes mathbfB = mathbfA_i, j cdot mathbfB_j, k = mathbfC_i, j, k$$

which means the $(i,j)^th$ vector of $mathbfA$ times the $(j, k)^th$ vector of $mathbfB$.

answered 17 mins ago

Daniel

1137

add a commentÂ |Â

up vote
0
down vote

Tensor multiplication is just a generalization of matrix multiplication which is just a generalization of vector multiplication.

Matrix multiplication is defined as:

$$ A_i cdot B_j = C_i, j$$

where $i$ is the $i^th$ row, $j$ is the $j^th$ column, and $cdot$ is the dot product. Therefore it just a series of dot products.

One can then see how this extends to tensors:
$$mathbfA_i cdot mathbfB_j = mathbfC_i, j $$

Assuming all tensors are of rank three(it can be described with three coordinates):

$$mathbfA otimes mathbfB = mathbfA_i, j cdot mathbfB_j, k = mathbfC_i, j, k$$

which means the $(i,j)^th$ vector of $mathbfA$ times the $(j, k)^th$ vector of $mathbfB$.

answered 17 mins ago

Daniel

1137

add a commentÂ |Â

up vote
0
down vote

Tensor multiplication is just a generalization of matrix multiplication which is just a generalization of vector multiplication.

Matrix multiplication is defined as:

$$ A_i cdot B_j = C_i, j$$

where $i$ is the $i^th$ row, $j$ is the $j^th$ column, and $cdot$ is the dot product. Therefore it just a series of dot products.

One can then see how this extends to tensors:
$$mathbfA_i cdot mathbfB_j = mathbfC_i, j $$

Assuming all tensors are of rank three(it can be described with three coordinates):

$$mathbfA otimes mathbfB = mathbfA_i, j cdot mathbfB_j, k = mathbfC_i, j, k$$

which means the $(i,j)^th$ vector of $mathbfA$ times the $(j, k)^th$ vector of $mathbfB$.

answered 17 mins ago

Daniel

1137

Tensor multiplication is just a generalization of matrix multiplication which is just a generalization of vector multiplication.

Matrix multiplication is defined as:

$$ A_i cdot B_j = C_i, j$$

where $i$ is the $i^th$ row, $j$ is the $j^th$ column, and $cdot$ is the dot product. Therefore it just a series of dot products.

One can then see how this extends to tensors:
$$mathbfA_i cdot mathbfB_j = mathbfC_i, j $$

Assuming all tensors are of rank three(it can be described with three coordinates):

$$mathbfA otimes mathbfB = mathbfA_i, j cdot mathbfB_j, k = mathbfC_i, j, k$$

which means the $(i,j)^th$ vector of $mathbfA$ times the $(j, k)^th$ vector of $mathbfB$.

answered 17 mins ago

Daniel

1137

answered 17 mins ago

Daniel

1137

answered 17 mins ago

Daniel

1137

answered 17 mins ago

Daniel

1137

add a commentÂ |Â

up vote
0
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answered 7 mins ago

Ahmad Bazzi

1011

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up vote
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answered 7 mins ago

Ahmad Bazzi

1011

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answered 7 mins ago

Ahmad Bazzi

1011

New contributor

answered 7 mins ago

Ahmad Bazzi

1011

New contributor

answered 7 mins ago

Ahmad Bazzi

1011

New contributor

answered 7 mins ago

Ahmad Bazzi

1011

answered 7 mins ago

Ahmad Bazzi

1011

New contributor

Ahmad Bazzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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