Mixing
Cross product of two sets with vectors as elements.
$begingroup$
If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?
I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and
$
mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
$
Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?
linear-algebra discrete-mathematics
edited Jan 25 at 5:15
Henno Brandsma
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
$endgroup$
add a comment |
$begingroup$
If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?
I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and
$
mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
$
Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?
linear-algebra discrete-mathematics
edited Jan 25 at 5:15
Henno Brandsma
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
$endgroup$
add a comment |
$begingroup$
If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?
I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and
$
mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
$
Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?
linear-algebra discrete-mathematics
edited Jan 25 at 5:15
Henno Brandsma
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
$endgroup$
If $X in mathcal{X} = {x_1,x_2}$, for some $x_1,x_2 in mathbb{R}^{n times 1}$ and $Y in mathcal{Y} = {y_1,y_2}$, for some $y_1,y_2 in mathbb{R}^{m times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $mathcal{X} times mathcal{Y}$ ?
I think not, because $Z = (X^T,Y^T)^T in mathcal{Z} = left { begin{pmatrix} x_1 \ y_1 end{pmatrix}, begin{pmatrix} x_1 \ y_2 end{pmatrix}, begin{pmatrix} x_2 \ y_1 end{pmatrix}, begin{pmatrix} x_2 \ y_2 end{pmatrix} right}$, and
$
mathcal{X} times mathcal{Y} = { (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) }.
$
Then, is there a way to write $mathcal{Z}$ in terms if $mathcal{X}$ and $mathcal{Y}$ ?
linear-algebra discrete-mathematics
linear-algebra discrete-mathematics
edited Jan 25 at 5:15
Henno Brandsma
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
edited Jan 25 at 5:15
Henno Brandsma
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
edited Jan 25 at 5:15
Henno Brandsma
113k348122
edited Jan 25 at 5:15
Henno Brandsma
113k348122
edited Jan 25 at 5:15
Henno Brandsma
113k348122
113k348122
asked Jan 25 at 5:09
OliVerOliVer
718
asked Jan 25 at 5:09
OliVerOliVer
718
asked Jan 25 at 5:09
OliVerOliVer
718
718
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$
for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).
So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as
$$(cdot)^T(A) = {x^T mid x in A}$$
so for readability you might as well define
$$A^T = (cdot)^T(A)$$
This means that you can define $mathcal{Z}$ as
$$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$
answered Jan 27 at 2:04
MetricMetric
1,25159
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$
for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).
So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as
$$(cdot)^T(A) = {x^T mid x in A}$$
so for readability you might as well define
$$A^T = (cdot)^T(A)$$
This means that you can define $mathcal{Z}$ as
$$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$
answered Jan 27 at 2:04
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$
for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).
So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as
$$(cdot)^T(A) = {x^T mid x in A}$$
so for readability you might as well define
$$A^T = (cdot)^T(A)$$
This means that you can define $mathcal{Z}$ as
$$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$
answered Jan 27 at 2:04
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$
for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).
So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as
$$(cdot)^T(A) = {x^T mid x in A}$$
so for readability you might as well define
$$A^T = (cdot)^T(A)$$
This means that you can define $mathcal{Z}$ as
$$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$
answered Jan 27 at 2:04
MetricMetric
1,25159
$endgroup$
Firstly, it seems that you've already defined the transpose $$(cdot)^T : bigcup_{n=1}^infty (mathbb{R}^{n}cupmathbb{R}^{n times 1}) rightarrow bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$$
for all row and column vectors (I'm just using plain $mathbb{R}^n$ for the "rows" rather than $mathbb{R}^{1 times n}$).
So the image of subsets $A subseteq bigcup_{n=1}^infty (mathbb{R}^{n}cup mathbb{R}^{n times 1})$ under $(cdot)^T$ are defined already as
$$(cdot)^T(A) = {x^T mid x in A}$$
so for readability you might as well define
$$A^T = (cdot)^T(A)$$
This means that you can define $mathcal{Z}$ as
$$mathcal{Z} = (mathcal{X}^T times mathcal{Y}^T)^T$$
answered Jan 27 at 2:04
MetricMetric
1,25159
answered Jan 27 at 2:04
MetricMetric
1,25159
answered Jan 27 at 2:04
MetricMetric
1,25159
answered Jan 27 at 2:04
MetricMetric
1,25159
1,25159
add a comment |
add a comment |
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