Jointly Gaussian and Brownian bridge

Question

let $X_t = B(t) - tB(1)$, for $t \in[0,1]$ the classical Brownian Bridge.

Let also $s \leq t$: why are $X_t$ and $X_t - \frac{1-t}{1-s} X_s$ jointly Gaussian?

I just know that $(X_t)$ is a Gaussian process and I'm able to show it: basically one takes a partition of $[0,1]$ and see that $(X_{t_1}, \ldots, X_{t_n})$ is the image via a linear mapping of $(B(t_1), \ldots, B(t_n), B(1))$.

Edit I think I found the answer by myself, but a check would br really appreciated.

So clearly the vector $(X_t - \frac{1-t}{1-s}X_s, X_s)$ is a linear map of the vector $(X_t,X_s)$. So the questions is: is the vector $(X_t,X_s)$ Gaussian?

Now, I look at each component:$(B(t) - tB(1), B(s)-sB(1))$. Again, this is a linear map of the vector $(B(s),B(t),B(1))$ which is a Gaussian vector because $(B_t)$ is a Brownian motion

This concludes the proof

Answer 1

If you multiply a Gaussian random vector by a constant matrix (i.e. a non-random matrix) then what you get is always a Gaussian random vector.

You have independent Gaussians $B_t$ and $B_1-B_t.$ Independent univariate Gaussians are jointly Gaussian. Your random vector is tuple of a constant linear combinations of these two.