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I know the rank of $\Sigma^{-1}m m'$ is $1$. How can I find the eigenvalue and eigenvector of it? ($\Sigma^{-1}$ is the inverse of a covariance matrix and $m$ is a mean vector.)

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Apply the definitions: a nonzero vector $x$ is an eigenvector of a matrix $\mathbb{A}$ means there exists a scalar $\lambda$ such that $\mathbb{A}x = \lambda x.$ (Eigenvectors are defined only up to nonzero scalar multiples.) When this is the case, $\lambda$ is the eigenvalue associated with (all nonzero multiples of) $x$. Whence, letting $\mathbb{A} = \Sigma^{-1}mm',$ we seek a vector $x$ for which

$$\Sigma^{-1}mm'x = \lambda x.$$

Notice that $m'x$ is a scalar. This means the left hand side is a scalar multiple of the vector $\Sigma^{-1}m$, which is nonzero provided $m$ is nonzero (an auxiliary assumption we will have to make). Consequently $x$ must be a multiple of $\Sigma^{-1}m.$ Let's check:

$$\left(\Sigma^{-1}mm'\right)(\Sigma^{-1}m) = \left(\Sigma^{-1}m\right)\left(m'\Sigma^{-1}m\right).$$

Sure enough, the right hand side is a multiple of $\Sigma^{-1}m$ and the multiple is explicitly given as $m'\Sigma^{-1}m$, which is the desired eigenvalue.

This analysis holds for any square matrix $\Sigma^{-1}$ and any vector $m$ for which $\Sigma^{-1}m$ is nonzero.

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