bilinear transformation $phi Utimes Vto W$ such that $Im(phi)={phi(u,v): uin U, vin V}$ is not a subspace of...
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Find a bilinear transformation $phi Utimes Vto W$ such that $Im(phi)={phi(u,v): uin U, vin V}$ is not a subspace of $W$
I truly don't have an idea otherwise to brute force lots of tries and find one that fits. Is there a technique of some sort that can help?
linear-algebra examples-counterexamples bilinear-form
edited 3 hours ago
Servaes
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
add a comment |
up vote
1
down vote
favorite
Find a bilinear transformation $phi Utimes Vto W$ such that $Im(phi)={phi(u,v): uin U, vin V}$ is not a subspace of $W$
I truly don't have an idea otherwise to brute force lots of tries and find one that fits. Is there a technique of some sort that can help?
linear-algebra examples-counterexamples bilinear-form
edited 3 hours ago
Servaes
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find a bilinear transformation $phi Utimes Vto W$ such that $Im(phi)={phi(u,v): uin U, vin V}$ is not a subspace of $W$
I truly don't have an idea otherwise to brute force lots of tries and find one that fits. Is there a technique of some sort that can help?
linear-algebra examples-counterexamples bilinear-form
edited 3 hours ago
Servaes
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
Find a bilinear transformation $phi Utimes Vto W$ such that $Im(phi)={phi(u,v): uin U, vin V}$ is not a subspace of $W$
I truly don't have an idea otherwise to brute force lots of tries and find one that fits. Is there a technique of some sort that can help?
linear-algebra examples-counterexamples bilinear-form
linear-algebra examples-counterexamples bilinear-form
edited 3 hours ago
Servaes
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
edited 3 hours ago
Servaes
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
edited 3 hours ago
Servaes
20.9k33789
edited 3 hours ago
Servaes
20.9k33789
edited 3 hours ago
Servaes
20.9k33789
20.9k33789
asked Nov 17 at 17:12
Guerlando OCs
6821450
asked Nov 17 at 17:12
Guerlando OCs
6821450
asked Nov 17 at 17:12
Guerlando OCs
6821450
6821450
To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33
add a comment |
To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33
To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33
To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33
add a comment |
1 Answer
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As far as I know there is no technique, but you might want to consider the case $U=V=Bbb{R}^2$ and the map $phi$ that sends a pair to the four coordinate products. That is to say
$$phi: Bbb{R}^2timesBbb{R}^2 longrightarrow Bbb{R}^4: ((x_1,y_1),(x_2,y_2)) longmapsto (x_1x_2,x_1y_2,y_1x_2,y_1y_2).$$
answered Nov 17 at 17:48
Servaes
20.9k33789
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
As far as I know there is no technique, but you might want to consider the case $U=V=Bbb{R}^2$ and the map $phi$ that sends a pair to the four coordinate products. That is to say
$$phi: Bbb{R}^2timesBbb{R}^2 longrightarrow Bbb{R}^4: ((x_1,y_1),(x_2,y_2)) longmapsto (x_1x_2,x_1y_2,y_1x_2,y_1y_2).$$
answered Nov 17 at 17:48
Servaes
20.9k33789
add a comment |
up vote
1
down vote
accepted
As far as I know there is no technique, but you might want to consider the case $U=V=Bbb{R}^2$ and the map $phi$ that sends a pair to the four coordinate products. That is to say
$$phi: Bbb{R}^2timesBbb{R}^2 longrightarrow Bbb{R}^4: ((x_1,y_1),(x_2,y_2)) longmapsto (x_1x_2,x_1y_2,y_1x_2,y_1y_2).$$
answered Nov 17 at 17:48
Servaes
20.9k33789
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As far as I know there is no technique, but you might want to consider the case $U=V=Bbb{R}^2$ and the map $phi$ that sends a pair to the four coordinate products. That is to say
$$phi: Bbb{R}^2timesBbb{R}^2 longrightarrow Bbb{R}^4: ((x_1,y_1),(x_2,y_2)) longmapsto (x_1x_2,x_1y_2,y_1x_2,y_1y_2).$$
answered Nov 17 at 17:48
Servaes
20.9k33789
As far as I know there is no technique, but you might want to consider the case $U=V=Bbb{R}^2$ and the map $phi$ that sends a pair to the four coordinate products. That is to say
$$phi: Bbb{R}^2timesBbb{R}^2 longrightarrow Bbb{R}^4: ((x_1,y_1),(x_2,y_2)) longmapsto (x_1x_2,x_1y_2,y_1x_2,y_1y_2).$$
answered Nov 17 at 17:48
Servaes
20.9k33789
answered Nov 17 at 17:48
Servaes
20.9k33789
answered Nov 17 at 17:48
Servaes
20.9k33789
answered Nov 17 at 17:48
Servaes
20.9k33789
20.9k33789
add a comment |
add a comment |
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To show that W is a subspace you want to show that $u$+c$v$ $in$ W for $u,vin W$ and a scalar c. In this case, we want $phi(u_1,v_1) + c phi(u_2,v_2) in $ W. Since $phi$ is bilinear, we would need $c phi(u_2,v_2)= phi(cu_2,v_2) = phi(u_2,cv_2)$. So I guess that would be a place to start. Find a $phi$ where that isn't true.
– Joel Pereira
Nov 17 at 17:33