# Show that all the characteristic roots of a dispersion matrix of a random variable are non-negative

Show that all the characteristic roots of a dispersion matrix of a random variable are non-negative.

$$\begin{vmatrix} \sigma_{11}-\lambda & \sigma_{12} & \cdots & \sigma_{1p}\\ \sigma_{21} & \sigma_{22}-\lambda & \cdots & \sigma_{2p} \\ \vdots & \vdots & & \vdots\\ \sigma_{p1} & \sigma_{p2} &\cdots & \sigma_{pp}-\lambda \end{vmatrix}=0$$ Hence how can I show that $\lambda$ is non-negative. Please help.

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Hint: what are the implications of the characteristic roots being non-negative? Try to find a way to utilize that $y'\Sigma y=y'E(X-\mu)(X-\mu)'y=Var(y'(X-\mu))$. –  MånsT Sep 7 '12 at 14:25
@MånsT:please explain the implication and how I proceed –  Argha Sep 7 '12 at 14:40
Same hint, different language: the characteristic roots of that matrix are all variances of linear combinations of components of the variable. –  whuber Sep 7 '12 at 14:40
@Ranabir: have a look at the definition of positive-semidefinite matrices –  MånsT Sep 7 '12 at 15:00
I would have given the same hint as MansT. I think if you look at the properties of positive semidefinite matrices you might find a theorem about them that will answer your question. –  Michael Chernick Sep 7 '12 at 16:15

$(\Sigma-\lambda I)y=0$ The characteristic Equation
$=>y'(\Sigma-\lambda I)y=0$ Pre-Multipling $y'$
$=>y'\Sigma y=y'\lambda y$
$=>\lambda=\frac{y'\Sigma y}{y'y}=\frac{Var(y'(X-\mu))}{\sum y_i^2} \geq 0$ Asuming $\sum y_i^2\neq 0$