do eigenvectors correspond to direction of maximum scaling?

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.

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 06:59
Hey @GregMartin can you point me to any prove for your statement that it is true only for a symmetric matrix? — Rishabh Agrahari, Jun 28 '19 at 12:37
And so, do we have any ultimate useful result for when A is not symmetric? — information_interchange, Mar 25 '20 at 05:25
Thanks, this was a great question that really helped me understand that the spectral radius is larger than or equal to the largest eigenvalue, with equality only occurring when the matrix is symmetric — information_interchange, Mar 28 '20 at 01:55

score 6 · Answer 1 · answered Nov 26 '20 at 22:23

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.

score 3 · Answer 2 · edited Jan 05 '19 at 14:59

3

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 05 '19 at 14:59

Bhavishya Desai

3

answered Oct 12 '15 at 21:55

user3527854

39

do eigenvectors correspond to direction of maximum scaling?

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