do eigenvectors correspond to direction of maximum scaling?












3














Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553




No other vector when acted by this matrix will get stretched as much
as this eigenvector.




Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$



The largest eigenvalue is 1.35 (approx.)



Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$



magnitude(v) = 1.0



magnitude(A*v) = 1.373



So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)



Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.










share|cite|improve this question




















  • 1




    The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
    – Greg Martin
    Sep 16 '15 at 6:59










  • Thank you for the answer Greg.
    – user1837245
    Sep 16 '15 at 19:58
















3














Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553




No other vector when acted by this matrix will get stretched as much
as this eigenvector.




Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$



The largest eigenvalue is 1.35 (approx.)



Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$



magnitude(v) = 1.0



magnitude(A*v) = 1.373



So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)



Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.










share|cite|improve this question




















  • 1




    The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
    – Greg Martin
    Sep 16 '15 at 6:59










  • Thank you for the answer Greg.
    – user1837245
    Sep 16 '15 at 19:58














3












3








3


3





Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553




No other vector when acted by this matrix will get stretched as much
as this eigenvector.




Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$



The largest eigenvalue is 1.35 (approx.)



Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$



magnitude(v) = 1.0



magnitude(A*v) = 1.373



So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)



Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.










share|cite|improve this question















Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553




No other vector when acted by this matrix will get stretched as much
as this eigenvector.




Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$



The largest eigenvalue is 1.35 (approx.)



Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$



magnitude(v) = 1.0



magnitude(A*v) = 1.373



So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)



Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.







eigenvalues-eigenvectors






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edited Apr 13 '17 at 12:19









Community

1




1










asked Sep 16 '15 at 6:03









user1837245user1837245

162




162








  • 1




    The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
    – Greg Martin
    Sep 16 '15 at 6:59










  • Thank you for the answer Greg.
    – user1837245
    Sep 16 '15 at 19:58














  • 1




    The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
    – Greg Martin
    Sep 16 '15 at 6:59










  • Thank you for the answer Greg.
    – user1837245
    Sep 16 '15 at 19:58








1




1




The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
– Greg Martin
Sep 16 '15 at 6:59




The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
– Greg Martin
Sep 16 '15 at 6:59












Thank you for the answer Greg.
– user1837245
Sep 16 '15 at 19:58




Thank you for the answer Greg.
– user1837245
Sep 16 '15 at 19:58










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No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)



Linear Transformations & Eigenvectors Explained






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    No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)



    Linear Transformations & Eigenvectors Explained






    share|cite|improve this answer




























      1














      No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)



      Linear Transformations & Eigenvectors Explained






      share|cite|improve this answer


























        1












        1








        1






        No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)



        Linear Transformations & Eigenvectors Explained






        share|cite|improve this answer














        No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)



        Linear Transformations & Eigenvectors Explained







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        edited 2 days ago









        Bhavishya Desai

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        answered Oct 12 '15 at 21:55









        user3527854user3527854

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