# Taking constant out of multivariate normal

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?

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.
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 $$BX \sim N(B\mu,B\Sigma B')$$.