# How do you eyeball the eigenvectors?

I've been playing around--for longer than I care to admit--on this amazing site that allows you to visualize matrix transformations, determinants, and eigenvectors in order to develop an intuition of how they work. I've gotten a great intuition of matrix transformations and determinants, but I'm having a really hard time with eigenvectors. I understand the computation just fine. I also understand the "words" explaining the direction concept just fine. I even see that when I draw a vector that aligns with an eigenvector, that vector's direction doesn't change after the transformation. However, I'm just not getting the visceral understanding of the positioning of the eigenvectors and how they relate to the transformed basis. My gut is telling me that the transformed basis vectors define some "maximal direction of shear" and this can be used to find the direction of the eigenvectors.

QUESTION: Is there a geometrical construction or (even better) a way to "eyeball" the eigenvectors when you know the transformed basis vectors? An intuitive explanation (along with a drawing) for the non-mathematician is appreciated. ELI5 plz

ASIDE: One thing that I have noticed is that when I keep $$j$$ fixed and increase the magnitude of $$i$$, one of the eigenvectors seems to be "pulled" towards the $$i$$. The other eigenvector always points in the direction of $$j$$. Thus, there seems to be some interplay between the relative magnitudes of the transformed basis vectors and the angle between the original basis and the transformed basis.

12/29/2022

I left $$j$$ unchanged. I systematically altered $$i$$ to look for patterns in how eigenvectors change w.r.t changes in $$i$$. One eigenvector always remained parallel to $$j$$, so I only examined the effect on the other.

I first rotated $$i$$ from $$0$$ to $$90$$ degrees and found that the eigenvector sweeps an arc of a circle from $$-90$$ to $$-45$$ degrees.

Next, I scaled $$i$$ by varying factors from $$0.5$$ to $$1000$$ and rotated it as before after each scaling. I found some interesting patterns that I'm still pondering on.

Perhaps this will stimulate some thoughts.

Plot of systematically rotating i from 0 to 90 degrees.

• Try this: youtube.com/watch?v=PFDu9oVAE-g Dec 28, 2022 at 7:37
• Having seen your graphics, you might be interested by this question of mine about families of eigenvalues of a pencil of matrices : math.stackexchange.com/q/2611435/305862 Dec 29, 2022 at 22:45
• Thanks, Jean Marie! This is very useful. Dec 29, 2022 at 23:55

Here is a way to "eyeball" eigenvectors and eigenvalues as well in the particular case of symmetric matrices.

$$\pmatrix{a&b\\b&c}\pmatrix{x\\y}=\pmatrix{x'\\y'} \tag{1}$$

It sticks to your "maximal direction of shear" intuition.

Let us consider the image of the unit circle under the transformation which is an ellipse $$\frak{E}$$.

The main result (see explanation below) is that the major and minor axes of ellipse $$\frak{E}$$ are directed by the eigenvectors of the matrix.

Moreover, one can also visualize the eigenvalues as being the values of the semi-axes $$a=\lambda_1, b=\lambda_2$$ of ellipse $$\frak{E}$$ which means that with respect to the orthonormal axes $$X-Y$$ defined by the eigenvectors, the equation of the ellipse is :

$$\frac{X^2}{\lambda_1^2}+\frac{Y^2}{\lambda_2^2}=1$$

Remark: Here, we have taken matrix :

$$\pmatrix{2&2\\2&5} \ \ \text{with eigenvalues} \ \ \lambda_1=6, \ \lambda_2=1$$

$$\text{with associated eigenvectors} \ \ \pmatrix{1\\2}, \pmatrix{-2\\1}$$

Explanation: Using the inverse relationship of (1) : $$\pmatrix{d&e\\e&f}\pmatrix{x'\\y'}=\pmatrix{x\\y}$$

constraint $$x^2+y^2=1$$ gives:

$$(dx'+ey')^2+(ex'+fy')^2=1$$

which is the equation of a conic curve and it is easy to prove that it is the equation of an ellipse.

• Isn't this rather a method of eyeballing singular vectors? Dec 28, 2022 at 18:05
• @John Madden You pinpoint an interesting point but the "rather" isn't exact. Plainly, singular vectors of $B$ enter into this frame because they are eigenvectors of symmetric matrix $B^TB$... Dec 28, 2022 at 18:11
• Thank you for this thoughtful response +1. I will wait 2 days for an answer that generalizes to non-symmetric matrices before accepting. You clearly show how the "maximum direction of shear" points in direction of the $(1, 2)$ eigenvector; and for the symmetric matrix, the other eigenvector is orthogonal to it. To close the loop, are you saying that we can "eyeball" the eigenvectors by first drawing an ellipse based on 3 facts: center of the ellipse is the origin, the basis vectors are two known points $(2, 2)$ and $(2, 5)$ on the ellipse, and one of the axes will always be unit length? Dec 28, 2022 at 18:22
• Not these vectors but the following unit vectors constituting the orthonormal basis $\tfrac{1}{\sqrt{5}}(1,2)$ and $\tfrac{1}{\sqrt{5}}(-2,1)$. Dec 28, 2022 at 18:33

The visualizer linked to in the question is not the most general; it only allows for real values in the linear transformation. Furthermore, it is misleading in a sense: the initial basis vectors are always being displayed as an orthonormal basis. Finally, another point to keep in mind is that eigenvectors are not unique up to a scalar.

Is there a geometrical construction or (even better) a way to "eyeball" the eigenvectors when you know the transformed basis vectors?

Is really a question about geometry, but in order to talk about geometry we need to introduce some additional structure, like an inner product. Because you ask for a geometrical intepretation of the eigenvectors, but eigenvectors are not unique up to a scalar, any geometric answer you come up with will really have to do with the eigenrays.

Consider the linear transformation $$L$$ which sends the vector $$\hat{e}_1$$ to $$2\hat{e}_1$$ and sends $$\hat{e}_2$$ to $$3\hat{e}_1.$$ In the usual basis, $$L=\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.$$

Now let $$a$$ and $$b$$ be arbitrary vectors, and consider the linear transformation $$R$$ which sends $$a$$ to $$2a,$$ and $$b$$ to $$3b.$$ In the basis of these eigenvectors $$R=\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.$$

So is there really anything fundamentally different between these two transformation? No. They are different transformations, of course, but a priori there is no reason to prefer one set of basis vectors to another.

Of course, if we introduce some additional structure, then we can ask extra questions. Are the vectors $$\hat{e_1}$$ and $$\hat{e_2}$$ orthogonal? Do they have length $$1?$$ What about $$a$$ and $$b?$$

Jean Marie's answer to this question takes advantage of the fact that we live in Euclidean space, so symmetric matrices are intimately connected to rotation matrices. Thus it seems like there is a natural way to understand the eigenvalues/vectors of these matrices. Which there is, of course, but it has less to do with the eigenvalues/vectors and more to do with the rotational geometry implicit in the display we have assumed.

• Thanks for your insights. If linear transformations represent rotations, stretching, and shearing of space then intuitively we should be able to guess where vectors in that space are only being stretched. Let's forget the rigor of math for a moment and liken linear transformations to daily things. For example, can we consider a "rectangle of Play-Doh", subjecting it to some linear transformation (other than rotation), and then asking where are the lines only getting stretched? Dec 29, 2022 at 22:21