If $sum_{i=1}^vu_iotimes v_i =0$ then $v_i=0$ for all $i=1,cdots, n$
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Let $U$, $V$ be $K$-vector spaces. Let $u_1,cdots, u_nin U$ be linear independent vectors and $v_1,cdots, v_nin V$ arbitrary vectors. Show that if $sum_{i=1}^vu_iotimes v_i =0$ then $v_i=0$ for all $i=1,cdots, n$
I can obviously show the converse, but it does not suffice. I tried to find the answer here but I couldn't. It seems simple but I don't know how to begin
linear-algebra
asked Nov 17 at 18:07
Guerlando OCs
6821450
add a comment |
up vote
0
down vote
favorite
Let $U$, $V$ be $K$-vector spaces. Let $u_1,cdots, u_nin U$ be linear independent vectors and $v_1,cdots, v_nin V$ arbitrary vectors. Show that if $sum_{i=1}^vu_iotimes v_i =0$ then $v_i=0$ for all $i=1,cdots, n$
I can obviously show the converse, but it does not suffice. I tried to find the answer here but I couldn't. It seems simple but I don't know how to begin
linear-algebra
asked Nov 17 at 18:07
Guerlando OCs
6821450
1
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $U$, $V$ be $K$-vector spaces. Let $u_1,cdots, u_nin U$ be linear independent vectors and $v_1,cdots, v_nin V$ arbitrary vectors. Show that if $sum_{i=1}^vu_iotimes v_i =0$ then $v_i=0$ for all $i=1,cdots, n$
I can obviously show the converse, but it does not suffice. I tried to find the answer here but I couldn't. It seems simple but I don't know how to begin
linear-algebra
asked Nov 17 at 18:07
Guerlando OCs
6821450
Let $U$, $V$ be $K$-vector spaces. Let $u_1,cdots, u_nin U$ be linear independent vectors and $v_1,cdots, v_nin V$ arbitrary vectors. Show that if $sum_{i=1}^vu_iotimes v_i =0$ then $v_i=0$ for all $i=1,cdots, n$
I can obviously show the converse, but it does not suffice. I tried to find the answer here but I couldn't. It seems simple but I don't know how to begin
linear-algebra
linear-algebra
asked Nov 17 at 18:07
Guerlando OCs
6821450
asked Nov 17 at 18:07
Guerlando OCs
6821450
asked Nov 17 at 18:07
Guerlando OCs
6821450
asked Nov 17 at 18:07
Guerlando OCs
6821450
asked Nov 17 at 18:07
Guerlando OCs
6821450
6821450
1
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35
add a comment |
1
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35
1
1
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $w_1, ldots, w_m$ be a basis of $V$. Then $u_i otimes w_j$ is linearly independent in
$U otimes V$. One can write
$$sum_{i=1}^{n} u_i otimes v_i = sum_{i=1}^{n} u_i otimes left( sum_{j=1}^{m}c^i_j w_jright) = sum_{i=1}^{n} sum_{j=1}^{m} c^i_j (u_i otimes w_j) = 0.$$
Clearly, this implies that all $c^i_j = 0$. Thus, $v_i = sum c^i_j w_j = 0$.
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $w_1, ldots, w_m$ be a basis of $V$. Then $u_i otimes w_j$ is linearly independent in
$U otimes V$. One can write
$$sum_{i=1}^{n} u_i otimes v_i = sum_{i=1}^{n} u_i otimes left( sum_{j=1}^{m}c^i_j w_jright) = sum_{i=1}^{n} sum_{j=1}^{m} c^i_j (u_i otimes w_j) = 0.$$
Clearly, this implies that all $c^i_j = 0$. Thus, $v_i = sum c^i_j w_j = 0$.
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
add a comment |
up vote
1
down vote
accepted
Let $w_1, ldots, w_m$ be a basis of $V$. Then $u_i otimes w_j$ is linearly independent in
$U otimes V$. One can write
$$sum_{i=1}^{n} u_i otimes v_i = sum_{i=1}^{n} u_i otimes left( sum_{j=1}^{m}c^i_j w_jright) = sum_{i=1}^{n} sum_{j=1}^{m} c^i_j (u_i otimes w_j) = 0.$$
Clearly, this implies that all $c^i_j = 0$. Thus, $v_i = sum c^i_j w_j = 0$.
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $w_1, ldots, w_m$ be a basis of $V$. Then $u_i otimes w_j$ is linearly independent in
$U otimes V$. One can write
$$sum_{i=1}^{n} u_i otimes v_i = sum_{i=1}^{n} u_i otimes left( sum_{j=1}^{m}c^i_j w_jright) = sum_{i=1}^{n} sum_{j=1}^{m} c^i_j (u_i otimes w_j) = 0.$$
Clearly, this implies that all $c^i_j = 0$. Thus, $v_i = sum c^i_j w_j = 0$.
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
Let $w_1, ldots, w_m$ be a basis of $V$. Then $u_i otimes w_j$ is linearly independent in
$U otimes V$. One can write
$$sum_{i=1}^{n} u_i otimes v_i = sum_{i=1}^{n} u_i otimes left( sum_{j=1}^{m}c^i_j w_jright) = sum_{i=1}^{n} sum_{j=1}^{m} c^i_j (u_i otimes w_j) = 0.$$
Clearly, this implies that all $c^i_j = 0$. Thus, $v_i = sum c^i_j w_j = 0$.
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
answered Nov 17 at 18:32
Jeffery Opoku-Mensah
3,517719
3,517719
add a comment |
add a comment |
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1
Should it be $sum_{i=1}^n$ rather than $sum_{i=1}^v$ in the title and text?
– Semiclassical
Nov 17 at 18:35