TITLE: do eigenvectors correspond to direction of maximum scaling?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: There is something called the matrix norm, $||A||$ which according to Linear Algebra and it's Applications by Strang is defined on $\mathbb R^{n\times n}$ to be $$||A||=\max_{x\neq0}\frac{||Ax||}{||x||},$$
where $||x||=\sqrt{x^Tx}$. From this, $||Ax||\le||A||||x||$. So the norm measures the "amplifying effect" of $A$ on a vector $x$. Equality happens at the optimal $x$. The value $||Ax||/||x||$ is always nonnegative. We square this to get
$$||A||^2=\max_{x\neq0}\frac{x^TA^TAx}{x^Tx}.$$
Matrix $A^TA$ is symmetric. This is known as the Rayleigh quotient, as it is well known that its maximum is attained when $x$ is the eigenvector corresponding to the largest eigenvalue. Then the maximum happens when $x$ is the eigenvector of $A^TA$ corresponding to the largest eigenvalue of that matrix.