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This is regarding article in this link which explains Kalman filter. In this article at one point in equation 4 author says

If we multiply every point in a distribution by a matrix A, then what happens to its covariance matrix Σ?

Well, it’s easy. I’ll just give you the identity:

How can derive this equation?

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We have $$ Cov(Ax) = \Bbb E((Ax)(Ax)^T) = \Bbb E(A(xx^T)A^T) = A\Bbb E(xx^T) A^T = A Cov(x)A^T $$

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    sorry im new to math. Why Ax is written as (Ax)(Ax)T – 2017-01-12
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    @Sirish it's not. However, for a column-vector of random variables $y$, we have $Cov(y) = \Bbb E(yy^T)$. When $y = Ax$, we end up with $(Ax)(Ax)^T$ instead. – 2017-01-12
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I have a small addition: The OP has not specified that $x$ is zero mean.

In general, ${\rm COV}[x]$ is defined as $\mathbb{E}[(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T]$. If $x$ is zero mean, i.e., $\mathbb{E}[x]=0$, Omnomnomnom has already given the answer. If it is not, it still works, like this:

$$\begin{align}{\rm COV}[A x] & = \mathbb{E}[(Ax - \mathbb{E}[Ax])(Ax - \mathbb{E}[Ax])^T] \\ & = \mathbb{E}[(Ax - A\mathbb{E}[x])(Ax - A\mathbb{E}[x])^T ]\\ & = \mathbb{E}[A(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T A^T ] \\ & = A \mathbb{E}[(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T ]A^T \\ & = A {\rm COV}[x] A^T \\ \end{align}$$