## When Do 2×2 Covariance Matrices Have Repeated Eigenvalues?

The calculation answering the question in the title is not that difficult, but I needed it recently, and so I thought I would add it to my blog.

Suppose we have a $2 \times 2$ covariance matrix $\displaystyle \Sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} \end{bmatrix}.$

By definition, the eigenvalues are the solutions to the characteristic equation $(\sigma_{11}-\lambda)(\sigma_{22} - \lambda) - \sigma^2_{12} = 0$. From the quadratic formula we obtain

$\displaystyle \lambda = \frac{\sigma_{11} + \sigma_{22} \pm \sqrt{(\sigma_{11} + \sigma_{22})^2 - 4(\sigma_{11} \sigma_{22} - \sigma^2_{12})}}{2}.$

Repeated eigenvalues occur precisely when the discriminant (the expression under the square root sign) is $0$.  Thus repeated eigenvalues occur when $\displaystyle (\sigma_{11} + \sigma_{22})^2 - 4 \sigma_{11} \sigma_{22} + 4 \sigma^2_{12} = 0$,

or, simplifying,

$\displaystyle \sigma^2_{11} + 2 \sigma_{11} \sigma_{22} + \sigma^2_{22} - 4 \sigma_{11} \sigma_{22} + 4 \sigma^2_{12} = 0 \\ \Rightarrow \sigma^2_{11} - 2 \sigma_{11} \sigma_{22} + \sigma^2_{22} + 4 \sigma^2_{12} = 0 \\ \Rightarrow (\sigma_{11} - \sigma_{22})^2 + 4 \sigma^2_{12} = 0.$

Thus $\Sigma$ has repeated eigenvalues precisely when the two random variables are uncorrelated and have the same variance; i.e., $\sigma_{12} = 0$ and $\sigma_{11} = \sigma_{22}$.

Your conclusion is true in general for $n \times n$ symmetric matrices from the fact that any symmetric matrix is diagonalizable. Hence, $A = U \Sigma U^{-1}$. If all the eigenvalues are equal to $\sigma$, then $A = U \sigma U^{-1} = \sigma U U^{-1} = \sigma I$.