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If $\mathbf {Z}$ is random vector and $A$ is a fixed matrix, could someone explain why $$\mathrm{cov}[A \mathbf {Z}]= A \mathrm{cov}[\mathbf {Z}]A^\top.$$

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For a random (column) vector $\mathbf Z$ with mean vector $\mathbf{m} = E[\mathbf{Z}]$, the covariance matrix is defined as $\operatorname{cov}(\mathbf{Z}) = E[(\mathbf{Z}-\mathbf{m})(\mathbf{Z}-\mathbf{m})^T]$. Thus, the covariance matrix of $A\mathbf{Z}$, whose mean vector is $A\mathbf{m}$, is given by $$\begin{align}\operatorname{cov}(A\mathbf{Z}) &= E[(A\mathbf{Z}-A\mathbf{m})(A\mathbf{Z}-A\mathbf{m})^T]\\ &= E[A(\mathbf{Z}-\mathbf{m})(\mathbf{Z}-\mathbf{m})^TA^T]\\ &= AE[(\mathbf{Z}-\mathbf{m})(\mathbf{Z}-\mathbf{m})^T]A^T\\ &= A\operatorname{cov}(\mathbf{Z})A^T. \end{align}$$

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    $\begingroup$ I corrected my typo. Thanks for pointing out my error. $\endgroup$
    – user92612
    Aug 29 '14 at 14:54
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I would add to the answer by Dilip Sarwate that the same result holds also for transformation of the form $\mathbf{Z}A^T$: $$\mathrm{cov}(\mathbf{Z}A^T) = A\mathrm{cov}(\mathbf{Z})A^T$$

Using the same approach: $$ \begin{align} \mathrm{cov}(\mathbf{Z}A^T)&=\mathbb{E}[(\mathbf{Z}A^T-\mathbf{m}A^T)(\mathbf{Z}A^T-\mathbf{m}A^T)^T] \\ &=\mathbb{E}[(\mathbf{Z}-\mathbf{m})A^TA(\mathbf{Z}-\mathbf{m})^T] \\ &=\mathbb{E}[A(\mathbf{Z}-\mathbf{m})(\mathbf{Z}-\mathbf{m})^TA^T] \\ &=A\mathbb{E}[(\mathbf{Z}-\mathbf{m})(\mathbf{Z}-\mathbf{m})^T]A^T \\ &=A\mathrm{cov}(\mathbf{Z})A^T \\ \end{align} $$

Using $AB^TBA^T=BAA^TB^T$ in step (3): $$\begin{align}AB^TBA^T &= \left(\left(AB^TBA^T\right)^T\right)^T \\ &= \left(BA^TAB^T\right)^T \\ &= B\left(BA^TA\right)^T \\ &= BAA^TB^T \end{align}$$

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  • $\begingroup$ Why is $AB^TBA^T=BAA^TB^T$? $\endgroup$ Feb 21 '20 at 17:42
  • $\begingroup$ Looking at it now, this step seems incorrect to me... Thanks for pointing that out! $\endgroup$ Feb 25 '20 at 11:17

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