do eigenvectors correspond to direction of maximum scaling?
$begingroup$
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
$endgroup$
add a comment |
$begingroup$
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
$endgroup$
1
$begingroup$
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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
$begingroup$
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
$endgroup$
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
eigenvalues-eigenvectors
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Sep 16 '15 at 6:03
user1837245user1837245
163
asked Sep 16 '15 at 6:03
user1837245user1837245
163
asked Sep 16 '15 at 6:03
user1837245user1837245
163
163
1
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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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
1
$begingroup$
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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
1
1
$begingroup$
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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
1 Answer
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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
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
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$begingroup$
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
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
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
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
edited Jan 5 at 14:59
Bhavishya Desai
32
edited Jan 5 at 14:59
Bhavishya Desai
32
edited Jan 5 at 14:59
Bhavishya Desai
32
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
answered Oct 12 '15 at 21:55
user3527854user3527854
111
answered Oct 12 '15 at 21:55
user3527854user3527854
111
111
add a comment |
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$begingroup$
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.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58