Subspace of a vector space problem
$begingroup$
So the task says:
for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?
I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:
$Ax + Cy ∈ U$,
I could not find the answer so i checked the answer.
I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.
Thank you!
linear-algebra vector-spaces
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
$endgroup$
add a comment |
$begingroup$
So the task says:
for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?
I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:
$Ax + Cy ∈ U$,
I could not find the answer so i checked the answer.
I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.
Thank you!
linear-algebra vector-spaces
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
$endgroup$
2
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35
add a comment |
$begingroup$
So the task says:
for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?
I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:
$Ax + Cy ∈ U$,
I could not find the answer so i checked the answer.
I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.
Thank you!
linear-algebra vector-spaces
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
$endgroup$
So the task says:
for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?
I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:
$Ax + Cy ∈ U$,
I could not find the answer so i checked the answer.
I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.
Thank you!
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
edited Jan 21 at 15:02
José Carlos Santos
163k22130233
163k22130233
asked Jan 20 at 19:30
PetarPetar
317
asked Jan 20 at 19:30
PetarPetar
317
asked Jan 20 at 19:30
PetarPetar
317
317
2
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35
add a comment |
2
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35
2
2
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
add a comment |
$begingroup$
Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
add a comment |
$begingroup$
Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
add a comment |
$begingroup$
Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:33
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
add a comment |
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:05
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
$endgroup$
– José Carlos Santos
Jan 20 at 21:34
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
Ok thank you! I understand it now.
$endgroup$
– Petar
Jan 20 at 21:51
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
If my answer was useful, perhaps that you could mark it as the accepted one.
$endgroup$
– José Carlos Santos
Jan 20 at 21:53
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
$begingroup$
yes i did it now ,tnx
$endgroup$
– Petar
Jan 20 at 22:09
add a comment |
$begingroup$
Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
$begingroup$
Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
$begingroup$
Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
$endgroup$
Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:41
Shubham JohriShubham Johri
5,177717
5,177717
add a comment |
add a comment |
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2
$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
Jan 20 at 19:34
$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
Jan 20 at 20:13
$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
Jan 20 at 20:35