Show that $d^T Z\sim N(d^T\mu, d^TVd)$

Question

Consider $Z=(Z_1,\ldots,Z_n)^T\sim N(\mu,V)$ with $\mu=(\mu_1,\ldots,\mu_n)^T$ and $V=\text{Cov}(Z)$. Show that for $d\in\mathbb{R}^n$ it is $$ d^TZ\sim N(d^T\mu,d^TVd). $$

For me it is not clear how to show that $d^TZ$ is multivariate normal distributed. If that would be the case then because of $$ E(d^TZ)=d^TE(Z)=d^T\mu $$ and $$ \text{Cov}(d^TZ)=d^T\text{Cov}(Z)(d^T)^T=d^T\text{Cov}(Z)d $$ everything would be shown.

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I've read that $d^TZ$ is multivariate normal distributed exactly then when for each $b\in\mathbb{R}$ $$ bd^TZ $$ is univariate normal distributed.

So I choose an arbitrary $b\in\mathbb{R}$ and would have to show that $$ bd^TZ=\sum_{i=1}^{n}bd_iZ_i~~~(*) $$ is univariate normal distributed. If the $Z_i$ were independent then it would be clear because a linear combination of independent univariate normal distributed random variables is itself univariate normal distributed.

But here the $Z_i$ are not independent, what to do now to show that (*) is univariate normal distributed?

Answer 1

My Suggestion:

$Var(d^TZ)=E(d^T(Z-\mu)(Z-\mu)^Td)=d^T\cdot E((Z-\mu)(Z-\mu)^T)\cdot d$