Proof: If $lambda$ is an eigenvalue of $A$ and $x$ is a corresponding eigenvector, then $sAx=slambda x$ for...
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 '18 at 22:25
Eevee Trainer
4,4471632
asked Nov 20 '18 at 22:11
noobcoder
134
add a comment |
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 '18 at 22:25
Eevee Trainer
4,4471632
asked Nov 20 '18 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 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 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 '18 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 '18 at 22:16
add a comment |
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 '18 at 22:25
Eevee Trainer
4,4471632
asked Nov 20 '18 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 '18 at 22:25
Eevee Trainer
4,4471632
asked Nov 20 '18 at 22:11
noobcoder
134
edited Nov 20 '18 at 22:25
Eevee Trainer
4,4471632
asked Nov 20 '18 at 22:11
noobcoder
134
edited Nov 20 '18 at 22:25
Eevee Trainer
4,4471632
edited Nov 20 '18 at 22:25
Eevee Trainer
4,4471632
edited Nov 20 '18 at 22:25
Eevee Trainer
4,4471632
4,4471632
asked Nov 20 '18 at 22:11
noobcoder
134
asked Nov 20 '18 at 22:11
noobcoder
134
asked Nov 20 '18 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 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 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 '18 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 '18 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 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 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 '18 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 '18 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 '18 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 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 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 '18 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 '18 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 '18 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 '18 at 22:16
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 Nov 20 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
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 Nov 20 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
add a comment |
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 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
add a comment |
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 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
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 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
answered Nov 20 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
answered Nov 20 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
answered Nov 20 '18 at 22:29
Mohammad Riazi-Kermani
40.8k42059
40.8k42059
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
Nov 20 '18 at 22:14
If you have two equals things, just apply the same operation to both sides
– Blumer
Nov 20 '18 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 '18 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 '18 at 22:16