# Meaning of Third Eigenvalue in a Tilted Ellipse in a 3x3 Matrix

• I
Summary:
I modeled an ellipse of the form x^2 + 2bx + 2cxy + 2dy + ey^2=1 with a 3x3 matrix and am unsure what the meaning of the third eigenvalue is, since there are only two axes.
While reading the Strang textbook on tilted ellipses in the form of ax^2+2bxy +cy^2=1, I got to thinking about ellipses of the form ax^2 + 2bx + 2cxy + 2dy + ey^2=1 and wondered if I could model them through 3x3 symmetric matrices. I think I figured out something that worked for xT A x, where x = (x, y, 1). For example, x^2+4x+6xy+8y+2y^2=1 has matrix A with row 1= 1,3,2; row 2= 3,2,4; row 3=2,4,0

Anyway, my question involves the meaning of the three eigenvalues. In the 2x2 case, they are the semi-major and semi-minor axes. What about in this 3x3 case? I would assume two of them are again related to the axes, but what about the third? Or perhaps I just made up a procedure that was non-sensical.

Also, I assume this is a linear algebra topic, but if it would be better somewhere else, please let me know.

anuttarasammyak
Gold Member
Say an ellipse is on xy plane, I guess z axis might be the third one of ellipsoid with eigenvalue zero.

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Say an ellipse is on xy plane, I guess z axis might be the third one of ellipsoid with eigenvalue zero.

I think you'd be right if my vector, x, was (x, y, z), but it's (x, y, 1). I was doing that to account for 2bx and a 2dy terms (whereas the 2 x 2 only has x^2, y^2, and xy terms). Perhaps this step isn't valid, but without my choice of this vector, I don't know how you'd model something like x^2 + 4x +6xy + 8y + 2y^2 with a positive definite matrix. For this example, I checked the eigenvalues and they are all non-zero (approximately -1.05, -3.15, 7.21).

pasmith
Homework Helper
You are starting with $f(x,y,z) = ax^2 + 2cxy + ey^2 + 2bxz + 2dyz$ and setting $z = 1$.

Whether the eigenvalues mean anything depends on how you construct your matrix. If you construct a symmetric 3x3 matrix they likely don't mean anything for your 2D ellipse. However, if you set $$A = \begin{pmatrix} a & c & 0 \\ c & e & 0 \\ 2b & 2d & 0 \end{pmatrix}$$ then $(0,0,1)$ is an eigenvector with eigenvalue 0. The other eigenvalues are then the eigenvalues of $$\begin{pmatrix} a & c \\ c & e \end{pmatrix}.$$ The components of the eigenvectors satisfy \begin{align*} ax + cy &= \lambda x \\ cx + ey &= \lambda y \end{align*} as for the 2x2 matrix, but also $$2(bx + dy) = \lambda z$$ which is the only equation involving $z$ and is immediately solved.

However it may be better to eliminate the linear terms by writing $$ax^2 + 2cxy + ey^2 + 2bx + 2dy = a(x - x_0)^2 + 2c(x - x_0)(y - y_0) + e(y - y_0)^2 - C$$ where \begin{align*} ax_0 + cy_0 &= -b \\ cx_0 + ey_0 &= -d \\ ax_0^2 + 2cx_0y_0 + ey_0^2 &= C \end{align*} and $$ax^2 + 2cxy + ey^2 + 2bx + 2dy = 1$$ becomes $$\frac{a}{1+ C}(x - x_0)^2 + \frac{2c}{1 + C}(x -x_0)(y - y_0) + \frac{e}{1 + C}(y - y_0)^2 = 1.$$ So the effect of the linear terms is to translate and scale the ellipse.

etotheipi