An eigenvector for a linear operator with matrix $A$, is a non-zero vector $\vec{v}$ such that:

for some $\lambda \in \mathbb{R}$.

This means that the matrix $A$ is scaling $\vec{v}$. This means stretching, shrinking, or sometimes even reflecting $\vec{v}$ to go in the opposite direction.

Find the right entries for the matrix $A$ such that $A$ scales every input $\vec{v}$ by $\lambda = 2.$ Follow the pictures and descriptions below.

The matrix $A = \begin{bmatrix} 1.17 & -0.76 \\ 0.76 & 1.17 \end{bmatrix}$ has no eigenvectors. See some examples for yourself below. In the last frame you can drag the vector $\vec{v}$ around and confirm. There is a fixed, non-zero angle between every vector and its image.

In this example $A$ is a diagonal matrix. The x-axis and y-axis are comprised of eigenvectors but nothing else.

The very first example you saw consisted of a matrix with the same values on the diagonal:

These are the only matrices where every non-zero vector is an eigenvector. For every other kind of matrix we either get eigen-axes or no eigenvectors at all. I would like to show you a visualization for each type, but first we need to tweak the way we see these vectors a little bit. Instead of every vector $\vec{v}$ coming from the origin, we will just represent $\vec{v}$ as a point. The image, $A\vec{v}$, will be a vector coming out of that point. This way we can look at lots of vector pairs at the same time.

There is one small but important caveat for the pictures above: The orange output vectors were not scaled properly. Proper scaling would have quickly lead to giant orange overlapping vectors as we travelled out from the origin, since outputs are proportional to inputs when in the same direction. Each orange vector was displayed as a unit vector version of its self. For example, the two eigen axes is photo #4 above have different eigenvalues, but their arrows are all the same size.