3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.