What is a basis for the space of multilinear maps from $V_1 times dots times V_k to W$?
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I know that a basis for the space $L(V_1, dots, V_k; mathbb R)$ is $${varepsilon^{i_1} otimes cdots otimes varepsilon^{i_k}mid 1 le varepsilon^{i_j} le dim(V_j)}$$
where $varepsilon^{i_j}$ is the dual basis vector to the basis vector $E_{i_j}$ for $V_j$.
But what if we change this to $L(V_1, dots, V_k; W)$. What would be a basis for this space?
linear-algebra abstract-algebra tensor-products tensors
asked Nov 16 at 23:57
Al Jebr
4,12843175
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show 1 more comment
up vote
0
down vote
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I know that a basis for the space $L(V_1, dots, V_k; mathbb R)$ is $${varepsilon^{i_1} otimes cdots otimes varepsilon^{i_k}mid 1 le varepsilon^{i_j} le dim(V_j)}$$
where $varepsilon^{i_j}$ is the dual basis vector to the basis vector $E_{i_j}$ for $V_j$.
But what if we change this to $L(V_1, dots, V_k; W)$. What would be a basis for this space?
linear-algebra abstract-algebra tensor-products tensors
asked Nov 16 at 23:57
Al Jebr
4,12843175
If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that a basis for the space $L(V_1, dots, V_k; mathbb R)$ is $${varepsilon^{i_1} otimes cdots otimes varepsilon^{i_k}mid 1 le varepsilon^{i_j} le dim(V_j)}$$
where $varepsilon^{i_j}$ is the dual basis vector to the basis vector $E_{i_j}$ for $V_j$.
But what if we change this to $L(V_1, dots, V_k; W)$. What would be a basis for this space?
linear-algebra abstract-algebra tensor-products tensors
asked Nov 16 at 23:57
Al Jebr
4,12843175
I know that a basis for the space $L(V_1, dots, V_k; mathbb R)$ is $${varepsilon^{i_1} otimes cdots otimes varepsilon^{i_k}mid 1 le varepsilon^{i_j} le dim(V_j)}$$
where $varepsilon^{i_j}$ is the dual basis vector to the basis vector $E_{i_j}$ for $V_j$.
But what if we change this to $L(V_1, dots, V_k; W)$. What would be a basis for this space?
linear-algebra abstract-algebra tensor-products tensors
linear-algebra abstract-algebra tensor-products tensors
asked Nov 16 at 23:57
Al Jebr
4,12843175
asked Nov 16 at 23:57
Al Jebr
4,12843175
asked Nov 16 at 23:57
Al Jebr
4,12843175
asked Nov 16 at 23:57
Al Jebr
4,12843175
asked Nov 16 at 23:57
Al Jebr
4,12843175
4,12843175
If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45
|
show 1 more comment
If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45
If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45
|
show 1 more comment
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If $W$ is $n$-dimensional, then $Wcong mathbb{R}^n$; maps into $mathbb{R}^n$ are in one-to-one correspondence with families of $n$ maps into $mathbb{R}$; so you get a direct product of $n$ copies of the space $L(V_1,ldots,_k;mathbb{R})$; and then you can use the general facts about how to get a basis for $Xtimes Y$ when you have a basis for $X$ and a basis for $Y$.
– Arturo Magidin
Nov 17 at 0:09
@ArturoMagidin So, $L(V_1, dots, V_k; mathbb R^n) cong prod_{i=1}^k L(V_1, dots, V_k; mathbb R)$?
– Al Jebr
Nov 17 at 1:00
Yes; this is true because of the universal property of the product. Careful if $W$ is infinite dimensinoal, though, because then it is isomorphic to the direct sum of copies of $mathbb{R}$, and maps into the direct sum are not in one-to-one correspondence with families of maps.
– Arturo Magidin
Nov 17 at 1:07
A natural basis to take is ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ where the $epsilon^j$ range over a basis for the dual of $V_j$, and the $w$ ranges over a basis for $W$. It's a very natural extension of what you already have: each $epsilon$ eats a vector from a $V$, and rather than getting a number out the other side, you get a vector of $W$.
– Joppy
Nov 17 at 1:24
@Joppy But how does ${epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j}$ act on vectors? The element $epsilon^{i_1} otimes cdots otimes epsilon^{i_k} otimes w_j$ takes in $k$ vectors, but there are $k+1$ positions in that tensor product.
– Al Jebr
Nov 18 at 17:45