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For a univariate normal distribution $X \sim N(0, k\sigma^2)$ we can take out the $k$ to get $\sqrt{k}X \sim N(0, \sigma^2)$. In the multivariate normal case is there something similar? If $\textbf{Y} \sim N(\textbf{0}, \textbf{C}\sigma^2)$, can I say that $\textbf{C}^{\frac{1}{2}}\textbf{Y} \sim N(\textbf{0}, \sigma^2)$? If yes, how do I prove it?

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You have some errors: Actually $\frac{1}{\sqrt{k}}X\sim N(0,\sigma^2)$, which corresponds to $\textbf{C}^{-1/2}Y\sim N(0,\sigma^2\textbf{I})$. In MV normal RVs, it's well known that $Y=PX$ yields $Y\sim N(P\mu_x,PC_xP^T)$; you can look here for its proof.

So, here, $\mathbf{C}^{-1/2}\mathbf{Y}\sim N(\mathbf{0},\mathbf{C^{-1/2}C}\sigma^2\mathbf{C^{-T/2}})$, where $\mathbf{C}^{-1/2}=\mathbf{C}^{-T/2}$ and $\mathbf{C^{-1/2}CC^{-1/2}}=\mathbf{I}$, assuming that $\mathbf{C}$ was invertible, in which it might not be. And, recall that it is already symmetric, since $\mathbf{C}\sigma^2$ is the covariance matrix of $\mathbf{Y}$.

Note: $\mathbf{C^{-1/2}}$ can be calculated via diagonalization if applicable.

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Actually, for the multivariate case, given a random variable $X \sim N(\mu,\Sigma )$, the linearity property of the normal distribution says that if you multiply $X$ by a $p\times p$ matrix $B$ then $BX \sim N(B\mu,B\Sigma B')$.

For a proof, check the second theorem in this note.

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