How to show that if we multiply an orthogonal matrix by an n dimensional brownian motion is still a brownian motion

Question

I am currently learning about brownian motion, and I am working on an exercice which it's goal is to show that if we multiply an orthogonal matrix by an n dimensional brownian motion is still a brownian motion.

I just wanna know if there is a problem with my answer.

So my answer is as what follows:

First I have shown that the components are independent, I have shown that by showing that the components are gaussian, for the independence I have used the fact for gaussian independence is equivalent to Uncorrelatedness.

Second, I have shown that each component is a brownian motion.

is this correct or I have missed something.

Answer 1

For $A\in O(n)$ and $B_t$ an $n$-dimensional Brownian motion, write $W_t=AB_t\,.$ Clearly, $W_t$ is a continuous martingale. From $$ W^i_t=\sum_{k=1}^nA_{ik}B^k_t\,,\quad\text{ and }\quad\langle B^k,B^l\rangle_t=\delta_{kl}\,t $$ it follows that $$ \langle W^i,W^j\rangle_t=\sum_{k=1}^n A_{ik}\,A_{jk}\,t=AA^\top\,t=t\,. $$ Therefore, $W$ is a Brownian motion.