show that a E is a subspace of V [closed]
$begingroup$
I got a little problem with a math problem here it is if you guys can help me it will be wonderful thanks a lot.
1) Let $V$ be a vector space, $f$ an endomorphism of $V$
2) Let $lambda$ be the eigenvalue of $f$.
3) Let $E$ be defined as $E = { v in V , f(v) = lambda * v} $.
Q: Show that $E$ is a subspace of $V$ .
linear-algebra eigenvalues-eigenvectors
edited Dec 12 '18 at 5:53
AnyAD
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
$endgroup$
closed as off-topic by anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh Dec 12 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I got a little problem with a math problem here it is if you guys can help me it will be wonderful thanks a lot.
1) Let $V$ be a vector space, $f$ an endomorphism of $V$
2) Let $lambda$ be the eigenvalue of $f$.
3) Let $E$ be defined as $E = { v in V , f(v) = lambda * v} $.
Q: Show that $E$ is a subspace of $V$ .
linear-algebra eigenvalues-eigenvectors
edited Dec 12 '18 at 5:53
AnyAD
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
$endgroup$
closed as off-topic by anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh Dec 12 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35
add a comment |
$begingroup$
I got a little problem with a math problem here it is if you guys can help me it will be wonderful thanks a lot.
1) Let $V$ be a vector space, $f$ an endomorphism of $V$
2) Let $lambda$ be the eigenvalue of $f$.
3) Let $E$ be defined as $E = { v in V , f(v) = lambda * v} $.
Q: Show that $E$ is a subspace of $V$ .
linear-algebra eigenvalues-eigenvectors
edited Dec 12 '18 at 5:53
AnyAD
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
$endgroup$
I got a little problem with a math problem here it is if you guys can help me it will be wonderful thanks a lot.
1) Let $V$ be a vector space, $f$ an endomorphism of $V$
2) Let $lambda$ be the eigenvalue of $f$.
3) Let $E$ be defined as $E = { v in V , f(v) = lambda * v} $.
Q: Show that $E$ is a subspace of $V$ .
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Dec 12 '18 at 5:53
AnyAD
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
edited Dec 12 '18 at 5:53
AnyAD
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
edited Dec 12 '18 at 5:53
AnyAD
2,108812
edited Dec 12 '18 at 5:53
AnyAD
2,108812
edited Dec 12 '18 at 5:53
AnyAD
2,108812
2,108812
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
asked Dec 12 '18 at 5:45
JoshuaKJoshuaK
61
61
closed as off-topic by anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh Dec 12 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh Dec 12 '18 at 8:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – anomaly, The Chaz 2.0, Trevor Gunn, copper.hat, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35
add a comment |
2
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35
2
2
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
multiplication by $lambda$ is an endomorphism of $E$, say $S_{lambda}$.
the endomorphisms of $E$ are a ring.
so $f - S_{lambda}$ is an endomorphism of E.
therefore $ker(f -S_{lambda})$ is a subspace of E.
but $v in ker(f -S_{lambda})$ iff $f(v) -S_{lambda(v)} = 0$, i.e. $f(v) = lambda v$
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
add a comment |
$begingroup$
Take two vectors in $E $ say $v_1,v_2$ and show $v_1+v_2in E$. For the scalar $alpha $ show $alpha v_1in E$.
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
$endgroup$
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
multiplication by $lambda$ is an endomorphism of $E$, say $S_{lambda}$.
the endomorphisms of $E$ are a ring.
so $f - S_{lambda}$ is an endomorphism of E.
therefore $ker(f -S_{lambda})$ is a subspace of E.
but $v in ker(f -S_{lambda})$ iff $f(v) -S_{lambda(v)} = 0$, i.e. $f(v) = lambda v$
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
add a comment |
$begingroup$
multiplication by $lambda$ is an endomorphism of $E$, say $S_{lambda}$.
the endomorphisms of $E$ are a ring.
so $f - S_{lambda}$ is an endomorphism of E.
therefore $ker(f -S_{lambda})$ is a subspace of E.
but $v in ker(f -S_{lambda})$ iff $f(v) -S_{lambda(v)} = 0$, i.e. $f(v) = lambda v$
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
add a comment |
$begingroup$
multiplication by $lambda$ is an endomorphism of $E$, say $S_{lambda}$.
the endomorphisms of $E$ are a ring.
so $f - S_{lambda}$ is an endomorphism of E.
therefore $ker(f -S_{lambda})$ is a subspace of E.
but $v in ker(f -S_{lambda})$ iff $f(v) -S_{lambda(v)} = 0$, i.e. $f(v) = lambda v$
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
$endgroup$
multiplication by $lambda$ is an endomorphism of $E$, say $S_{lambda}$.
the endomorphisms of $E$ are a ring.
so $f - S_{lambda}$ is an endomorphism of E.
therefore $ker(f -S_{lambda})$ is a subspace of E.
but $v in ker(f -S_{lambda})$ iff $f(v) -S_{lambda(v)} = 0$, i.e. $f(v) = lambda v$
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
answered Dec 12 '18 at 5:59
David HoldenDavid Holden
14.8k21224
14.8k21224
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
add a comment |
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
$begingroup$
hey david tks for ur answer but i didn t see in class what a ring is , i m gonna do it the way AnyAD wrote it
$endgroup$
– JoshuaK
Dec 12 '18 at 14:21
add a comment |
$begingroup$
Take two vectors in $E $ say $v_1,v_2$ and show $v_1+v_2in E$. For the scalar $alpha $ show $alpha v_1in E$.
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
$endgroup$
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
add a comment |
$begingroup$
Take two vectors in $E $ say $v_1,v_2$ and show $v_1+v_2in E$. For the scalar $alpha $ show $alpha v_1in E$.
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
$endgroup$
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
add a comment |
$begingroup$
Take two vectors in $E $ say $v_1,v_2$ and show $v_1+v_2in E$. For the scalar $alpha $ show $alpha v_1in E$.
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
$endgroup$
Take two vectors in $E $ say $v_1,v_2$ and show $v_1+v_2in E$. For the scalar $alpha $ show $alpha v_1in E$.
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
answered Dec 12 '18 at 5:57
AnyADAnyAD
2,108812
2,108812
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
add a comment |
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
Hey AnyAD , i don't have to show that 0 is in E too ?
$endgroup$
– JoshuaK
Dec 12 '18 at 14:22
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
That's my question how can i show that 1/ 0 is in E ( for this one i know ) if i can prove that the scalar av1 is in E then the scalar can be 0 2/ v1 + v2 in E ( this is what i can't prove ) 3/ Lamba * v1 in E ( same problem )
$endgroup$
– JoshuaK
Dec 12 '18 at 14:49
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
$begingroup$
@JoshuaK $E$ is a subset of $V$ (given). So you just need to verufy that it satusfies the subspace condutions: it is non-empty, is closed under vector addition, it is closed under scalar multiplication.
$endgroup$
– AnyAD
Dec 12 '18 at 21:11
add a comment |
2
$begingroup$
What are you having difficulty with? This is just an application of the definition of linearity and subspace.
$endgroup$
– copper.hat
Dec 12 '18 at 5:52
$begingroup$
with the scalar part i don't know how to prove it , For the scalar α show αv1 ∈ E.
$endgroup$
– JoshuaK
Dec 12 '18 at 15:35