Proof: If $lambda$ is an eigenvalue of $A$ and $x$ is a corresponding eigenvector, then $sAx=slambda x$ for...
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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
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Eevee Trainer
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noobcoder
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up vote
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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 yesterday
Eevee Trainer
97111
asked yesterday
noobcoder
134
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
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
yesterday
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
yesterday
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 yesterday
Eevee Trainer
97111
asked yesterday
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 yesterday
Eevee Trainer
97111
asked yesterday
noobcoder
134
edited yesterday
Eevee Trainer
97111
asked yesterday
noobcoder
134
edited yesterday
Eevee Trainer
97111
edited yesterday
Eevee Trainer
97111
edited yesterday
Eevee Trainer
97111
97111
asked yesterday
noobcoder
134
asked yesterday
noobcoder
134
asked yesterday
noobcoder
134
134
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
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
yesterday
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
yesterday
add a comment |
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
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
yesterday
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
yesterday
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
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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 yesterday
Mohammad Riazi-Kermani
40.2k41958
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 yesterday
Mohammad Riazi-Kermani
40.2k41958
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 yesterday
Mohammad Riazi-Kermani
40.2k41958
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 yesterday
Mohammad Riazi-Kermani
40.2k41958
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 yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
add a comment |
add a comment |
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You don't have to consider $0$, positive and negative values separately. For any scalar $s$, $sAx=s(Ax)$.
– John Douma
yesterday
If you have two equals things, just apply the same operation to both sides
– Blumer
yesterday
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
yesterday
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
yesterday