Why is a matrix multiplied by an eigenvector not parallel to that eigenvector?Eigenvalue and Eigenvector for the change of Basis MatrixUnder what conditions does a complex matrix with a real eigenvalue have a corresponding real eigenvector?An eigenvector is a non-zero vector such that…Proof that eigenvector corresponding to simple eigenvalue is continuousPlease help with the rank-1 matrix eigenvectoreigenvector equivalence for subspace of matrix?If every non-zero vectors be the eigenvector of a real matrix $A$, prove that $A$ is the scalar matrix $lambda I_n$.Finding a vector that is not an eigenvector of a matrixEigenvector and adjoint eigenvector not orthogonalProve that any non zero linear combination of two eigenvectors is also an eigenvector
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Why is a matrix multiplied by an eigenvector not parallel to that eigenvector?
Eigenvalue and Eigenvector for the change of Basis MatrixUnder what conditions does a complex matrix with a real eigenvalue have a corresponding real eigenvector?An eigenvector is a non-zero vector such that…Proof that eigenvector corresponding to simple eigenvalue is continuousPlease help with the rank-1 matrix eigenvectoreigenvector equivalence for subspace of matrix?If every non-zero vectors be the eigenvector of a real matrix $A$, prove that $A$ is the scalar matrix $lambda I_n$.Finding a vector that is not an eigenvector of a matrixEigenvector and adjoint eigenvector not orthogonalProve that any non zero linear combination of two eigenvectors is also an eigenvector
$begingroup$
If $lambda$ is a non-zero eigenvalue with a corresponding eigenvector $v$, then $A v$ is parallel to $v$.
This statement is false. Why is that? Would it be parallel to $lambda v$?
linear-algebra eigenfunctions
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
$endgroup$
|
show 1 more comment
$begingroup$
If $lambda$ is a non-zero eigenvalue with a corresponding eigenvector $v$, then $A v$ is parallel to $v$.
This statement is false. Why is that? Would it be parallel to $lambda v$?
linear-algebra eigenfunctions
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
$endgroup$
2
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
1
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
2
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
2
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55
|
show 1 more comment
$begingroup$
If $lambda$ is a non-zero eigenvalue with a corresponding eigenvector $v$, then $A v$ is parallel to $v$.
This statement is false. Why is that? Would it be parallel to $lambda v$?
linear-algebra eigenfunctions
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
$endgroup$
If $lambda$ is a non-zero eigenvalue with a corresponding eigenvector $v$, then $A v$ is parallel to $v$.
This statement is false. Why is that? Would it be parallel to $lambda v$?
linear-algebra eigenfunctions
linear-algebra eigenfunctions
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 8:51
Rodrigo de Azevedo
13k41960
13k41960
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
asked Mar 11 at 2:20
Samurai BaleSamurai Bale
645
645
2
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
1
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
2
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
2
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55
|
show 1 more comment
2
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
1
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
2
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
2
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55
2
2
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
1
1
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
2
2
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
2
2
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
It's hard to know what to make of your question as you didn't specify what matrix $vecv$ is an eigenvector of.
If $A$ were the matrix corresponding to that eigenvector, then $Avecv=lambdavecv$, by definition, meaning the statement would be true. Since you're sure the statement is false, then we can assume that $A$ is not the matrix corresponding to that eigenvector. In that case, of course, $Avecv$ need not equal $lambda vecv$.
If you're confused about what it means to be parallel or what an Eigenvector is, I recommend 3b1b's video on the topic.
answered Mar 11 at 8:45
PatrickPatrick
1018
$endgroup$
add a comment |
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$begingroup$
It's hard to know what to make of your question as you didn't specify what matrix $vecv$ is an eigenvector of.
If $A$ were the matrix corresponding to that eigenvector, then $Avecv=lambdavecv$, by definition, meaning the statement would be true. Since you're sure the statement is false, then we can assume that $A$ is not the matrix corresponding to that eigenvector. In that case, of course, $Avecv$ need not equal $lambda vecv$.
If you're confused about what it means to be parallel or what an Eigenvector is, I recommend 3b1b's video on the topic.
answered Mar 11 at 8:45
PatrickPatrick
1018
$endgroup$
add a comment |
$begingroup$
It's hard to know what to make of your question as you didn't specify what matrix $vecv$ is an eigenvector of.
If $A$ were the matrix corresponding to that eigenvector, then $Avecv=lambdavecv$, by definition, meaning the statement would be true. Since you're sure the statement is false, then we can assume that $A$ is not the matrix corresponding to that eigenvector. In that case, of course, $Avecv$ need not equal $lambda vecv$.
If you're confused about what it means to be parallel or what an Eigenvector is, I recommend 3b1b's video on the topic.
answered Mar 11 at 8:45
PatrickPatrick
1018
$endgroup$
add a comment |
$begingroup$
It's hard to know what to make of your question as you didn't specify what matrix $vecv$ is an eigenvector of.
If $A$ were the matrix corresponding to that eigenvector, then $Avecv=lambdavecv$, by definition, meaning the statement would be true. Since you're sure the statement is false, then we can assume that $A$ is not the matrix corresponding to that eigenvector. In that case, of course, $Avecv$ need not equal $lambda vecv$.
If you're confused about what it means to be parallel or what an Eigenvector is, I recommend 3b1b's video on the topic.
answered Mar 11 at 8:45
PatrickPatrick
1018
$endgroup$
It's hard to know what to make of your question as you didn't specify what matrix $vecv$ is an eigenvector of.
If $A$ were the matrix corresponding to that eigenvector, then $Avecv=lambdavecv$, by definition, meaning the statement would be true. Since you're sure the statement is false, then we can assume that $A$ is not the matrix corresponding to that eigenvector. In that case, of course, $Avecv$ need not equal $lambda vecv$.
If you're confused about what it means to be parallel or what an Eigenvector is, I recommend 3b1b's video on the topic.
answered Mar 11 at 8:45
PatrickPatrick
1018
answered Mar 11 at 8:45
PatrickPatrick
1018
answered Mar 11 at 8:45
PatrickPatrick
1018
answered Mar 11 at 8:45
PatrickPatrick
1018
1018
add a comment |
add a comment |
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2
$begingroup$
What is your definition of parallel?
$endgroup$
– Brian Fitzpatrick
Mar 11 at 2:23
1
$begingroup$
What are your definitions of eigenvector and eigenvalue?
$endgroup$
– Brian Borchers
Mar 11 at 2:27
$begingroup$
If $vecvin S_lambda$ $Avecvcdotvecv=lambdavecvcdotvecv=lambda||vecv||^2=1 iff ||vecv||^2=dfrac1lambda$
$endgroup$
– Sebastian Cor
Mar 11 at 3:05
2
$begingroup$
What makes you think the statement is false, Samurai? Can you show us a counterexample?
$endgroup$
– Gerry Myerson
Mar 11 at 4:05
2
$begingroup$
The only explantions that I can come up with are: (1) it's a calculation mistake, (2) perhaps $lambda=0$ and in someone's definition, two vectors are said to be parallel only if they are scalar multiples of each other (this is just a wild guess; personally, I treat $0$ as a vector that is parallel to every vector).
$endgroup$
– user1551
Mar 11 at 8:55