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Matrix free linear map decomposition
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$begingroup$
Consider $a_1,dots,a_ninmathbb{R}^n$ and identify $a_jinmathcal{L}(mathbb{R},mathbb{R}^n)$ via $varphimapsto varphi1$.
Also, consider $Ainmathcal{L}(mathbb{R}^n)$ given by
$$Acolon (x_1,dots,x_n)mapsto a_1x_1 + dots + a_nx_ntag{$star$}$$
What's the name or symbol of the map
$$mathcal{L}(mathbb{R},mathbb{R}^n)timesdotstimesmathcal{L}(mathbb{R},mathbb{R}^n)tomathcal{L}(mathbb{R}^n,mathbb{R}^n),quad(a_1,dots,a_n)mapsto A$$
where $A$ and $a_1,dots,a_n$ are related as in $(star)$?
I'd like to write e.g. $A=a_1otimesdotsotimes a_n$.
Is there some higher level concept that induces that map? (E.g. some time ago I wondered if this map would be the tensor product).
The matrix equivalent would be saying that the columns of $A$ are the column vectors $a_1,dots,a_n$ and writing $A=begin{bmatrix}a_1& dots &a_nend{bmatrix}$. However, I'd like to keep things matrix free.
Thanks in advance.
linear-algebra matrices coordinate-systems
asked Mar 17 at 15:09
RamenRamen
507412
$endgroup$
add a comment |
$begingroup$
Consider $a_1,dots,a_ninmathbb{R}^n$ and identify $a_jinmathcal{L}(mathbb{R},mathbb{R}^n)$ via $varphimapsto varphi1$.
Also, consider $Ainmathcal{L}(mathbb{R}^n)$ given by
$$Acolon (x_1,dots,x_n)mapsto a_1x_1 + dots + a_nx_ntag{$star$}$$
What's the name or symbol of the map
$$mathcal{L}(mathbb{R},mathbb{R}^n)timesdotstimesmathcal{L}(mathbb{R},mathbb{R}^n)tomathcal{L}(mathbb{R}^n,mathbb{R}^n),quad(a_1,dots,a_n)mapsto A$$
where $A$ and $a_1,dots,a_n$ are related as in $(star)$?
I'd like to write e.g. $A=a_1otimesdotsotimes a_n$.
Is there some higher level concept that induces that map? (E.g. some time ago I wondered if this map would be the tensor product).
The matrix equivalent would be saying that the columns of $A$ are the column vectors $a_1,dots,a_n$ and writing $A=begin{bmatrix}a_1& dots &a_nend{bmatrix}$. However, I'd like to keep things matrix free.
Thanks in advance.
linear-algebra matrices coordinate-systems
asked Mar 17 at 15:09
RamenRamen
507412
$endgroup$
$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
$endgroup$
– eepperly16
Mar 17 at 16:42
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
$endgroup$
– Ramen
Mar 17 at 17:09
add a comment |
$begingroup$
Consider $a_1,dots,a_ninmathbb{R}^n$ and identify $a_jinmathcal{L}(mathbb{R},mathbb{R}^n)$ via $varphimapsto varphi1$.
Also, consider $Ainmathcal{L}(mathbb{R}^n)$ given by
$$Acolon (x_1,dots,x_n)mapsto a_1x_1 + dots + a_nx_ntag{$star$}$$
What's the name or symbol of the map
$$mathcal{L}(mathbb{R},mathbb{R}^n)timesdotstimesmathcal{L}(mathbb{R},mathbb{R}^n)tomathcal{L}(mathbb{R}^n,mathbb{R}^n),quad(a_1,dots,a_n)mapsto A$$
where $A$ and $a_1,dots,a_n$ are related as in $(star)$?
I'd like to write e.g. $A=a_1otimesdotsotimes a_n$.
Is there some higher level concept that induces that map? (E.g. some time ago I wondered if this map would be the tensor product).
The matrix equivalent would be saying that the columns of $A$ are the column vectors $a_1,dots,a_n$ and writing $A=begin{bmatrix}a_1& dots &a_nend{bmatrix}$. However, I'd like to keep things matrix free.
Thanks in advance.
linear-algebra matrices coordinate-systems
asked Mar 17 at 15:09
RamenRamen
507412
$endgroup$
Consider $a_1,dots,a_ninmathbb{R}^n$ and identify $a_jinmathcal{L}(mathbb{R},mathbb{R}^n)$ via $varphimapsto varphi1$.
Also, consider $Ainmathcal{L}(mathbb{R}^n)$ given by
$$Acolon (x_1,dots,x_n)mapsto a_1x_1 + dots + a_nx_ntag{$star$}$$
What's the name or symbol of the map
$$mathcal{L}(mathbb{R},mathbb{R}^n)timesdotstimesmathcal{L}(mathbb{R},mathbb{R}^n)tomathcal{L}(mathbb{R}^n,mathbb{R}^n),quad(a_1,dots,a_n)mapsto A$$
where $A$ and $a_1,dots,a_n$ are related as in $(star)$?
I'd like to write e.g. $A=a_1otimesdotsotimes a_n$.
Is there some higher level concept that induces that map? (E.g. some time ago I wondered if this map would be the tensor product).
The matrix equivalent would be saying that the columns of $A$ are the column vectors $a_1,dots,a_n$ and writing $A=begin{bmatrix}a_1& dots &a_nend{bmatrix}$. However, I'd like to keep things matrix free.
Thanks in advance.
linear-algebra matrices coordinate-systems
linear-algebra matrices coordinate-systems
asked Mar 17 at 15:09
RamenRamen
507412
asked Mar 17 at 15:09
RamenRamen
507412
edited Mar 17 at 18:35
Ramen
asked Mar 17 at 15:09
RamenRamen
507412
asked Mar 17 at 15:09
RamenRamen
507412
asked Mar 17 at 15:09
RamenRamen
507412
507412
$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
$endgroup$
– eepperly16
Mar 17 at 16:42
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
$endgroup$
– Ramen
Mar 17 at 17:09
add a comment |
$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
$endgroup$
– eepperly16
Mar 17 at 16:42
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
$endgroup$
– Ramen
Mar 17 at 17:09
$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
$endgroup$
– eepperly16
Mar 17 at 16:42
$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
$endgroup$
– eepperly16
Mar 17 at 16:42
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
$endgroup$
– Ramen
Mar 17 at 17:09
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
$endgroup$
– Ramen
Mar 17 at 17:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I understand your question correctly, you are looking for an n-linear functional such that $(x_1, . . ., x_n) to sum_{i=1}^{n}a_{i}x_i$. Note that the scalar product of two arbitrary vectors $u, v$ is the following:
$$<u, v> = u_1v_1 + ... + u_nv_n = sum_{i=1}^{n}u_iv_i$$
So, for the map that you mentioned, we could define $M$ as the row vector with elements being the column vectors $a_i$ so that the scalar product $<M, X> = a_1x_1 + ... + a_nx_n.$
In other words, you are just looking for the scalar product operator.
answered Mar 17 at 16:37
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
add a comment |
$begingroup$
I am not sure I understand what you are asking, but maybe you are looking for the map from a vector space to its dual, which we could write as $a mapsto langle a, cdot rangle$. In general this depends on a metric $langle cdot,cdotrangle$ (or some other preferred non-degenerate bilinear form). In $R^n$ it is often implicitly assumed that you use the Euclidean one and, with respect to the canonical basis, what you are doing is, for $ain R^n$
$$amapsto alphain (R^n)^*=mathcal{L}(R^n,R),$$ where, for any $vin R^n$, $$alpha (v) =a^T v.$$
answered Mar 17 at 18:12
GFRGFR
3,3241023
$endgroup$
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I understand your question correctly, you are looking for an n-linear functional such that $(x_1, . . ., x_n) to sum_{i=1}^{n}a_{i}x_i$. Note that the scalar product of two arbitrary vectors $u, v$ is the following:
$$<u, v> = u_1v_1 + ... + u_nv_n = sum_{i=1}^{n}u_iv_i$$
So, for the map that you mentioned, we could define $M$ as the row vector with elements being the column vectors $a_i$ so that the scalar product $<M, X> = a_1x_1 + ... + a_nx_n.$
In other words, you are just looking for the scalar product operator.
answered Mar 17 at 16:37
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
add a comment |
$begingroup$
If I understand your question correctly, you are looking for an n-linear functional such that $(x_1, . . ., x_n) to sum_{i=1}^{n}a_{i}x_i$. Note that the scalar product of two arbitrary vectors $u, v$ is the following:
$$<u, v> = u_1v_1 + ... + u_nv_n = sum_{i=1}^{n}u_iv_i$$
So, for the map that you mentioned, we could define $M$ as the row vector with elements being the column vectors $a_i$ so that the scalar product $<M, X> = a_1x_1 + ... + a_nx_n.$
In other words, you are just looking for the scalar product operator.
answered Mar 17 at 16:37
Victoria MVictoria M
42618
$endgroup$
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
add a comment |
$begingroup$
If I understand your question correctly, you are looking for an n-linear functional such that $(x_1, . . ., x_n) to sum_{i=1}^{n}a_{i}x_i$. Note that the scalar product of two arbitrary vectors $u, v$ is the following:
$$<u, v> = u_1v_1 + ... + u_nv_n = sum_{i=1}^{n}u_iv_i$$
So, for the map that you mentioned, we could define $M$ as the row vector with elements being the column vectors $a_i$ so that the scalar product $<M, X> = a_1x_1 + ... + a_nx_n.$
In other words, you are just looking for the scalar product operator.
answered Mar 17 at 16:37
Victoria MVictoria M
42618
$endgroup$
If I understand your question correctly, you are looking for an n-linear functional such that $(x_1, . . ., x_n) to sum_{i=1}^{n}a_{i}x_i$. Note that the scalar product of two arbitrary vectors $u, v$ is the following:
$$<u, v> = u_1v_1 + ... + u_nv_n = sum_{i=1}^{n}u_iv_i$$
So, for the map that you mentioned, we could define $M$ as the row vector with elements being the column vectors $a_i$ so that the scalar product $<M, X> = a_1x_1 + ... + a_nx_n.$
In other words, you are just looking for the scalar product operator.
answered Mar 17 at 16:37
Victoria MVictoria M
42618
answered Mar 17 at 16:37
Victoria MVictoria M
42618
answered Mar 17 at 16:37
Victoria MVictoria M
42618
answered Mar 17 at 16:37
Victoria MVictoria M
42618
42618
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
add a comment |
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
$begingroup$
Thanks for your answer. I'm rather looking for the linear map $mathbb{R}^ntimesmathbb{R}^ntomathcal{L}(mathbb{R}^2,mathbb{R}^n)$ that maps vectors to linear maps. Anyway, I think now that my question was worded poorly, sorry!
$endgroup$
– Ramen
Mar 17 at 17:06
add a comment |
$begingroup$
I am not sure I understand what you are asking, but maybe you are looking for the map from a vector space to its dual, which we could write as $a mapsto langle a, cdot rangle$. In general this depends on a metric $langle cdot,cdotrangle$ (or some other preferred non-degenerate bilinear form). In $R^n$ it is often implicitly assumed that you use the Euclidean one and, with respect to the canonical basis, what you are doing is, for $ain R^n$
$$amapsto alphain (R^n)^*=mathcal{L}(R^n,R),$$ where, for any $vin R^n$, $$alpha (v) =a^T v.$$
answered Mar 17 at 18:12
GFRGFR
3,3241023
$endgroup$
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
add a comment |
$begingroup$
I am not sure I understand what you are asking, but maybe you are looking for the map from a vector space to its dual, which we could write as $a mapsto langle a, cdot rangle$. In general this depends on a metric $langle cdot,cdotrangle$ (or some other preferred non-degenerate bilinear form). In $R^n$ it is often implicitly assumed that you use the Euclidean one and, with respect to the canonical basis, what you are doing is, for $ain R^n$
$$amapsto alphain (R^n)^*=mathcal{L}(R^n,R),$$ where, for any $vin R^n$, $$alpha (v) =a^T v.$$
answered Mar 17 at 18:12
GFRGFR
3,3241023
$endgroup$
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
add a comment |
$begingroup$
I am not sure I understand what you are asking, but maybe you are looking for the map from a vector space to its dual, which we could write as $a mapsto langle a, cdot rangle$. In general this depends on a metric $langle cdot,cdotrangle$ (or some other preferred non-degenerate bilinear form). In $R^n$ it is often implicitly assumed that you use the Euclidean one and, with respect to the canonical basis, what you are doing is, for $ain R^n$
$$amapsto alphain (R^n)^*=mathcal{L}(R^n,R),$$ where, for any $vin R^n$, $$alpha (v) =a^T v.$$
answered Mar 17 at 18:12
GFRGFR
3,3241023
$endgroup$
I am not sure I understand what you are asking, but maybe you are looking for the map from a vector space to its dual, which we could write as $a mapsto langle a, cdot rangle$. In general this depends on a metric $langle cdot,cdotrangle$ (or some other preferred non-degenerate bilinear form). In $R^n$ it is often implicitly assumed that you use the Euclidean one and, with respect to the canonical basis, what you are doing is, for $ain R^n$
$$amapsto alphain (R^n)^*=mathcal{L}(R^n,R),$$ where, for any $vin R^n$, $$alpha (v) =a^T v.$$
answered Mar 17 at 18:12
GFRGFR
3,3241023
answered Mar 17 at 18:12
GFRGFR
3,3241023
answered Mar 17 at 18:12
GFRGFR
3,3241023
answered Mar 17 at 18:12
GFRGFR
3,3241023
3,3241023
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
add a comment |
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
$begingroup$
Thanks for your answer, however, this is still not what I intended, as it is analogous to the row decomposition of a matrix, right? If we identify vectors $a_1,dots,a_n$ as linear forms in the sense of Riesz, we can write $A=(a_1,dots,a_n)inmathcal{L}(mathbb{R}^n)$. I am interested in a similar notation for the column decomposition, but in terms of linear maps only. In particular, the mathematical machinery is not what I am really interested in (the thing is kind of obvious) but rather the name, or the right symbol, perhaps stemming from some higher level concept.
$endgroup$
– Ramen
Mar 17 at 18:27
add a comment |
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$begingroup$
While you may want to keep your notation cordinate free, your operation is inherently depend on the coordinates in the domain as you map the standard basis vectors $e_k$ to specific vectors $a_k$.
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– eepperly16
Mar 17 at 16:42
$begingroup$
@eepperly16 Yes, you're right, thank you! (The choice of basis happened by choosing the isomorphism to be isometric, i.e. choosing $varphimapstovarphi 1$.) Please see my edit.
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– Ramen
Mar 17 at 17:09