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Is it necessary that the individual random variables to be drawn from a Gaussian distribution to be considered jointly Gaussian?

I consider a symmetric positive definite Covariance matrix be considered as a test for Joint Gaussianity in case of bivariate random variables, I am I right?

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    If $X=(X_1,\ldots,X_n)$ is gaussian then yes, each $X_k$ is gaussian. Is this your question? – 2017-01-11
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    Is it always true? So if one variable is non-Gaussian and other is Gaussian, there is no way there joint distribution could be Gaussian? – 2017-01-11
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    Yes it is always true. Have you some kind of definition of gaussianity of a random vector at your disposal? This could prove useful... – 2017-01-11

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The components of a multivariate Gaussian random variable are each [one-dimensional] Gaussian. However, the converse is not necessarily true; see here for some discussion and examples.

I am not sure what your last sentence means. All multivariate random variables have positive semidefinite covariance matrices, so that is not a way to characterize multivariate Gaussian distributions.

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    The last sentence means if covariance matrix is strictly Positive definite , will that be a condition to identify joint Gaussian , since in the expression for Joint Gaussian distribution a A covariance matrix inverse is present – 2017-01-11
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    @JonNy Then my point stands. For example, if $X$ and $Y$ are independent random variables, then the covariance matrix of $(X,Y)$ is diagonal with diagonal elements $Var(X)$ and $Var(Y)$, so the covariance matrix can be strictly positive definite, even though $(X,Y)$ is not Gaussian. – 2017-01-11