tensor rank of an element in a tensor product
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0
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Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.
How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.
abstract-algebra commutative-algebra tensor-products tensor-rank
asked Nov 22 at 14:38
idriskameni
608
add a comment |
up vote
0
down vote
favorite
Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.
How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.
abstract-algebra commutative-algebra tensor-products tensor-rank
asked Nov 22 at 14:38
idriskameni
608
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.
How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.
abstract-algebra commutative-algebra tensor-products tensor-rank
asked Nov 22 at 14:38
idriskameni
608
Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.
How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.
abstract-algebra commutative-algebra tensor-products tensor-rank
abstract-algebra commutative-algebra tensor-products tensor-rank
asked Nov 22 at 14:38
idriskameni
608
asked Nov 22 at 14:38
idriskameni
608
asked Nov 22 at 14:38
idriskameni
608
asked Nov 22 at 14:38
idriskameni
608
asked Nov 22 at 14:38
idriskameni
608
608
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Suppose rank $t=r$ and write
$$
t = sum_{i=1}^r v_i otimes w_i
$$
where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.
Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
Then we have
$$
t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
$$
contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.
answered Nov 22 at 14:53
Eric
2088
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
add a comment |
up vote
1
down vote
Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.
answered Nov 22 at 15:32
Joppy
5,573420
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
up vote
2
down vote
accepted
Suppose rank $t=r$ and write
$$
t = sum_{i=1}^r v_i otimes w_i
$$
where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.
Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
Then we have
$$
t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
$$
contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.
answered Nov 22 at 14:53
Eric
2088
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
add a comment |
up vote
2
down vote
accepted
Suppose rank $t=r$ and write
$$
t = sum_{i=1}^r v_i otimes w_i
$$
where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.
Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
Then we have
$$
t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
$$
contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.
answered Nov 22 at 14:53
Eric
2088
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose rank $t=r$ and write
$$
t = sum_{i=1}^r v_i otimes w_i
$$
where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.
Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
Then we have
$$
t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
$$
contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.
answered Nov 22 at 14:53
Eric
2088
Suppose rank $t=r$ and write
$$
t = sum_{i=1}^r v_i otimes w_i
$$
where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.
Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
Then we have
$$
t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
$$
contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.
answered Nov 22 at 14:53
Eric
2088
answered Nov 22 at 14:53
Eric
2088
answered Nov 22 at 14:53
Eric
2088
answered Nov 22 at 14:53
Eric
2088
2088
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
add a comment |
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
– idriskameni
Dec 4 at 12:04
1
1
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
– Eric
Dec 4 at 13:01
add a comment |
up vote
1
down vote
Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.
answered Nov 22 at 15:32
Joppy
5,573420
add a comment |
up vote
1
down vote
Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.
answered Nov 22 at 15:32
Joppy
5,573420
add a comment |
up vote
1
down vote
up vote
1
down vote
Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.
answered Nov 22 at 15:32
Joppy
5,573420
Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.
answered Nov 22 at 15:32
Joppy
5,573420
answered Nov 22 at 15:32
Joppy
5,573420
answered Nov 22 at 15:32
Joppy
5,573420
answered Nov 22 at 15:32
Joppy
5,573420
5,573420
add a comment |
add a comment |
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