# Does the covariance of i.i.d. random vectors/multivariate random variables have any zero terms?

If we have i.i.d. random variables, $$X$$ and $$Y$$, then $$\text{Cov}(X,Y)=0$$.

But let's say we have i.i.d. random vectors $$\boldsymbol{X}$$ and $$\boldsymbol{Y}$$, where $$\boldsymbol{X}=(X_{1},...,X_{p})$$ and $$\boldsymbol{Y}=(Y_{1},...,Y_{q})$$. Do we have any properties analogous to the univariate case: $$\text{Cov}(X,Y)=0$$, that always hold true, as a result of these random vectors being i.i.d.? Something like $$\text{Cov}(X_i,X_j)=0$$ if $$i\neq j$$, or $$\text{Cov}(X_i,Y_i)=0$$, or maybe the independence of $$\boldsymbol{X}$$ and $$\boldsymbol{Y}$$ gives a covariance matrix with zero terms everywhere except the diagonal (these are all just guesses, and not necessarily true). Are there any properties like this for i.i.d. random vectors?

If the random vectors $$X, Y$$ are independent, then, as functions of independent random variables are independent (see Functions of Independent Random Variables), $$X_i$$ and $$Y_j$$ are independent, since selecting one given component from a vector is a function. But the covariance of independent random variables are zero, that gives you the result.
In your question you speculate about a condition $$i \not= j$$, but that is unnecessary here. $$X_i$$ and $$Y_i$$ re also independent. The covariance matrix $$\DeclareMathOperator{\C}{\mathbb{C}} \C(x,y)$$ will be the zero matrix, also the diagonal.