From Introduction to Probability here (PDF page 348, book page 337) we are told:
Definition 7.5.1
A $k$-dimensional random vector $X=(X_1, ..., X_k)$ is Multivariate Normal if every linear combination of the $X_j$ has a Normal distribution. That is, we require
$$t_1X_1 + \cdots + t_kX_k$$
to have a Normal distribution for any constants $t_1, ..., t_k$.
Question set-up
Logically speaking this is of the form: $Q$ if $P$, where:
$P$: We have a $k$-dimensional random vector $X=(X_1, ..., X_k)$ where every linear combination of the $X_j$ has a Normal distribution
$Q$: $X=(X_1, ..., X_k)$ is Multivariate Normal.
So, if $P$ then $Q$ i.e. $P \implies Q$.
Question
If we are told that we have a MVN (i.e. a Bivariate Normal, $k=2$) with random vector $(X, Y)$ can we conclude that every linear combination of the $X, Y$ are Normal? It seems to me we cannot as we are given $Q$ is true, not $P$.
If Definition 7.5.1 above said "if and only if" then I think we could, but not just with "if"?
Note: I know that it does happen to be the case that every linear combination of the $X, Y$ is Normal if $(X, Y)$ are MVN...I'm asking if we can logically deduce it from the information given.
Background
This is from homework question 3 b (page 20 of PDF). Provided below for context:
Let $(X, Y)$ be Bivariate Normal, with $X$ and $Y$ marginally $\mathcal{N}(0, 1)$ and with correlation $\rho$ between $X$ and $Y$.
Show that $(X + Y,X − Y )$ is also Bivariate Normal.
Solution given: the linear combination $s(X + Y ) + t(X − Y ) = (s + t)X + (s − t)Y$ is also a linear combination of $X$ and $Y$, so it is Normal, which shows that $(X +Y,X −Y )$ is MVN.
