Proof: If $lambda$ is an eigenvalue of $A$ and $x$ is a corresponding eigenvector, then $sAx=slambda x$ for...
up vote
0
down vote
favorite
It's fairly straight forward to multiple $s$ on both sides of $Ax=lambda x$, for $x$ is not the $0$ vector. How do you go about showing that $sAx=slambda x$ holds true for every $s$ scalar?
I'm thinking of 3 scenarios, when $s=0, s<0, and s>0$.
When $s=0, 0Ax = 0lambda x$, this is valid since $x$ is not the $0$ vector to hold true.
When $s>0, sAx = slambda x$, which is scalar multiplication and is valid.
When $s<0, -sAx = -slambda x$ is equivalent to $sAx = slambda x$.
Is this enough to show the above statement is true for every scalar $s$?
Thank you
linear-algebra proof-writing eigenvalues-eigenvectors
edited Nov 20 at 22:25
Eevee Trainer
2,179220
asked Nov 20 at 22:11
noobcoder
134
add a comment |
up vote
0
down vote
favorite
It's fairly straight forward to multiple $s$ on both sides of $Ax=lambda x$, for $x$ is not the $0$ vector. How do you go about showing that $sAx=slambda x$ holds true for every $s$ scalar?
I'm thinking of 3 scenarios, when $s=0, s<0, and s>0$.
When $s=0, 0Ax = 0lambda x$, this is valid since $x$ is not the $0$ vector to hold true.
When $s>0, sAx = slambda x$, which is scalar multiplication and is valid.
When $s<0, -sAx = -slambda x$ is equivalent to $sAx = slambda x$.
Is this enough to show the above statement is true for every scalar $s$?
Thank you
linear-algebra proof-writing eigenvalues-eigenvectors
edited Nov 20 at 22:25
Eevee Trainer
2,179220
asked Nov 20 at 22:11
noobcoder
134
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
1
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It's fairly straight forward to multiple $s$ on both sides of $Ax=lambda x$, for $x$ is not the $0$ vector. How do you go about showing that $sAx=slambda x$ holds true for every $s$ scalar?
I'm thinking of 3 scenarios, when $s=0, s<0, and s>0$.
When $s=0, 0Ax = 0lambda x$, this is valid since $x$ is not the $0$ vector to hold true.
When $s>0, sAx = slambda x$, which is scalar multiplication and is valid.
When $s<0, -sAx = -slambda x$ is equivalent to $sAx = slambda x$.
Is this enough to show the above statement is true for every scalar $s$?
Thank you
linear-algebra proof-writing eigenvalues-eigenvectors
edited Nov 20 at 22:25
Eevee Trainer
2,179220
asked Nov 20 at 22:11
noobcoder
134
It's fairly straight forward to multiple $s$ on both sides of $Ax=lambda x$, for $x$ is not the $0$ vector. How do you go about showing that $sAx=slambda x$ holds true for every $s$ scalar?
I'm thinking of 3 scenarios, when $s=0, s<0, and s>0$.
When $s=0, 0Ax = 0lambda x$, this is valid since $x$ is not the $0$ vector to hold true.
When $s>0, sAx = slambda x$, which is scalar multiplication and is valid.
When $s<0, -sAx = -slambda x$ is equivalent to $sAx = slambda x$.
Is this enough to show the above statement is true for every scalar $s$?
Thank you
linear-algebra proof-writing eigenvalues-eigenvectors
linear-algebra proof-writing eigenvalues-eigenvectors
edited Nov 20 at 22:25
Eevee Trainer
2,179220
asked Nov 20 at 22:11
noobcoder
134
edited Nov 20 at 22:25
Eevee Trainer
2,179220
asked Nov 20 at 22:11
noobcoder
134
edited Nov 20 at 22:25
Eevee Trainer
2,179220
edited Nov 20 at 22:25
Eevee Trainer
2,179220
edited Nov 20 at 22:25
Eevee Trainer
2,179220
2,179220
asked Nov 20 at 22:11
noobcoder
134
asked Nov 20 at 22:11
noobcoder
134
asked Nov 20 at 22:11
noobcoder
134
134
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
1
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16
add a comment |
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
1
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
1
1
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Since $lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = lambda x$$
Note that $Ax$ and $lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=slambda x$
More interesting is the fact that $$A(sx)=lambda (sx) $$ which is part of the proof for eigen space generated by $x$
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since $lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = lambda x$$
Note that $Ax$ and $lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=slambda x$
More interesting is the fact that $$A(sx)=lambda (sx) $$ which is part of the proof for eigen space generated by $x$
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
Since $lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = lambda x$$
Note that $Ax$ and $lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=slambda x$
More interesting is the fact that $$A(sx)=lambda (sx) $$ which is part of the proof for eigen space generated by $x$
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = lambda x$$
Note that $Ax$ and $lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=slambda x$
More interesting is the fact that $$A(sx)=lambda (sx) $$ which is part of the proof for eigen space generated by $x$
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
Since $lambda$ is an eigenvalue with corresponding eigenvector $x$ we have $$Ax = lambda x$$
Note that $Ax$ and $lambda x$ are identical vectors and scalar multiplication is a well defined operator therefore for every scalar $s$ you get $sAx=slambda x$
More interesting is the fact that $$A(sx)=lambda (sx) $$ which is part of the proof for eigen space generated by $x$
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
answered Nov 20 at 22:29
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006972%2fproof-if-lambda-is-an-eigenvalue-of-a-and-x-is-a-corresponding-eigenvect%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
Nov 20 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 at 22:14
Doesn't this just follow from the fact that $Ax = lambda x$ by the definition of eigenvalues/eigenvectors, and then you could just multiply on the left by the scalar without a problem since they're equivalent?
– Eevee Trainer
Nov 20 at 22:14
1
This is trivial the way you wrote the title, as you suggested (gollows from propertues of matrix multiplication). The question likely is to show that if $x $ is an eigenvector corresponding to eigenvalue $lambda $ then $sx $ is also an eigenvector of $A$ correspond to the same eigenvalue (for non-zero $s $).
– AnyAD
Nov 20 at 22:16