Let $(B_{t})_{t \in [0,1]}$ be a brownian motion. Show that $B_{t}-tB_{1}$ is independent of $B_{1}$
My idea: I need to find a linear transformation $A$ that renders $A \begin{pmatrix} B_{1}-B_{t} \\ B_{t} \end{pmatrix}= \begin{pmatrix} B_{t}-tB_{1} \\ B_{1} \end{pmatrix}$
I feel like this is impossible since $A$ is a linear transformation and can thus not contain $B_{t}$ or $B_{1}$
Since the linear combinations of $V:=(B_t-tB_1,B_1)$ are Gaussian, $V$ is a Gaussian vector. It thus suffices to look at the covariance between the two components of $V$. Since $t\leqslant 1$, $\operatorname{Cov}\left(B_t,B_1\right)=t$ and $\operatorname{Cov}\left(-tB_1,B_1\right)=-t\operatorname{Cov}\left( B_1,B_1\right)=-t$.