Eigenvalues and Eigenvectors of $2 \times 2$ Matrix
Let's say I have a $2 \times 2$ matrix (actually the structure tensor of a discrete image - I): $$ \begin{bmatrix} \frac{\partial I}{\partial x}\frac{\partial I}{\partial x} & \frac{\partial I}{\partial x}\frac{\partial I}{\partial y} \\\ \frac{\partial I}{\partial y}\frac{\partial I}{\partial x} & \frac{\partial I}{\partial y}\frac{\partial I}{\partial y} \end{bmatrix}$$
It has 2 properties:
- Symmetric.
- Positive Semidefinite.
Given those properties, what would be the easiest method to numerically compute its eigenvectors (orthogonal) and eigenvalues?
Solution 1:
As a two-by-two matrix, applying Jacobi's method in fact gives the answer at once!
It is known that given a two-by-two symmetric matrix $\mathbf A$, one can construct an orthogonal matrix $\mathbf V=\bigl(\begin{smallmatrix}c&s\\\\-s&c\end{smallmatrix}\bigr)$ such that $\mathbf V^\top \mathbf A\mathbf V$ is diagonal, where the two numbers $c$ and $s$ satisfy $c^2+s^2=1$.
If the off-diagonal elements are not zero (why?), computing $c$ and $s$ can be done like so:
$$\begin{align*} \tau&=\frac{a_{22}-a_{11}}{2a_{12}}\\ t&=\frac{\mathrm{sgn}(\tau)}{|\tau|+\sqrt{1+\tau^2}}\\ c&=\frac1{\sqrt{1+t^2}}\\ s&=ct \end{align*}$$
from which the eigendecomposition easily follows.
(See Golub and Van Loan's excellent book for further details.)
Here is a Mathematica demonstration of Jacobi's method:
a = N[{{3, -1}, {-1, 2}}, 20];
tau = (a[[2, 2]] - a[[1, 1]])/a[[1, 2]]/2;
t = Sign[tau]/(Abs[tau] + Sqrt[1 + tau^2]);
{c, s} = {1, t}/Sqrt[1 + t^2]
{0.85065080835203993218, 0.5257311121191336060}
{l1, l2} = {a[[1, 1]] - t a[[1, 2]], a[[2, 2]] + t a[[1, 2]]}
{3.6180339887498948482, 1.3819660112501051518}
Eigenvalues[a] == {l1, l2}
True
Eigenvectors[a] == {{c, -s}, {s, c}}
True
a.{c, -s} == l1 {c, -s}
True
a.{s, c} == l2 {s, c}
True
Solution 2:
Since you work with a $2\times2$ matrix, the corresponding characteristic polynomial is quadratic so the eigenvalues can be expressed in closed form in terms of the matrix elements. As soon as you get the eigenvalues all you have to do is to solve two systems of linear equations with two unknowns (i.e. the coordinates of the corresponding eigenvector in the standard basis). Again, everything is nice and explicit, no need to apply a fancy numerical method.
Solution 3:
Despite other answers, I thought it might benefit the impatient to see the explicit answer below.
Let $$M = \left( \begin{array}{cc} a & b \\ b & c \end{array} \right),$$ be the input matrix.
Define the discriminant: $\Delta = \sqrt{a^2+4 b^2-2 a c+c^2}$
Then, the eigenvalues of $M$ are given by:
$\lambda_1 = 0.5(a+c-\Delta)$ and $\lambda_2 = 0.5(a+c+\Delta)$
Now, you can find a matrix $V$ such that $$M = V^{-1} \begin{pmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{pmatrix}V.$$
Mathematica says that the matrix $V$ is given by $$ V = \begin{pmatrix} \frac{a-c-\Delta}{2b} & 1\\ \frac{a-c+\Delta}{2b} & 1 \end{pmatrix} $$
If you are looking for orthogonal $V$, then the above calculations need some changes.