How is eigen vector of biggest eigen value is the most stretching direction?

Question

Here's citation from Linear Algebra and its Applications

If $\lambda_1$ is the eigen value with the greatest magnitude, then a corresponding unit eigenvector $v_1$ identifies a direction in which the stretching effect of $A$ is greatest. That is, the length of $Ax$ is maximized when $x = v_1$.

And I found an example that shows that the biggest eigen vector is not the most stretching direction (using this application)

This is a sphere transformed by a matrix and the matrix's 3 eigen vectors in cyan, purple, yellow. And more the sphere's surface is on the span of eigen vectors, more the color of surface resembles the eigen vector's color. The purple eigen vector is the biggest one with eigen value of 2.63 but, to be the most stretched direction, it should tilt a bit left and be more like red one isn't it?

So what is the book talking about exactly? Is it simply saying "The biggest eigen vector is the direction in which the transformation stretches most compared to other eigen vectors" something very obvious like this?

Answer 1

The claim is true for the maximum singular value of any matrix. So this is always true:

If $\sigma_1$ is the singular value with the greatest magnitude, then a corresponding unit right singular vector $v_1$ identifies a direction in which the stretching effect of $A$ is greatest. That is, the length of $Ax$ is maximized when $x=v_1$. Furthermore the direction becomes $u_1$, that is $Av_1=\sigma_1 u_1$ where $u_1$ is the corresponding left singular vector with unit length.

The thing is if $A$ is symmetric then $u_1=v_1$ and $\sigma_1=|\lambda_1|$.

Answer 2

It was a stupid question. I missed the part "when the matrix is symmetric". I tried symmetric matrix in the application and it demonstrated the statement very clearly: