do eigenvectors correspond to direction of maximum scaling?

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.

• 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