Let $X=(X_1,...,X_n)$ be a random column vector. Suppose $E[X]=\mu$ and $\operatorname{Cov}(X)=\Sigma$.
Show that $\operatorname{Cov}(\beta + BX)=B\Sigma B^T$, where $\beta$ is a k dimensional vector and $B$ is a $k \times n$ matrix.
Thoughts: So I'm not quite sure how to go about. At first glance somehting about quadratic forms and symmetric matrix, which I believe $\sigma$ comes to mind but I'm not sure if and how I can use it... I suppose I could try a calculate the matrix product on the right but this seems like a very tedious way of showing it. So any suggestions and hints would help!
The covariance matrix of a random vector $\vec{X}=\left(X_{1},\dots,X_{n}\right)^{T}$ is by definition: $$\mathbb{E}\left(\vec{X}-\mathbb{E}\vec{X}\right)\left(\vec{X}-\mathbb{E}\vec{X}\right)^{T}$$
Observe for random vector $\vec{Y}=\left(Y_{1},\dots,Y_{k}\right)^{T}$ that: $$\vec{Y}-\mathbb{E}\vec{Y}=\vec{\beta}+B\vec{X}-\mathbb{E}\left[\vec{\beta}+B\vec{X}\right]=\vec{\beta}+B\vec{X}-\vec{\beta}-B\mathbb{E}\vec{X}=B\left(\vec{X}-\mathbb{E}\vec{X}\right)$$
So that for the covariance matrix of $\vec{Y}$ we find:$$\mathbb{E}\left(\vec{Y}-\mathbb{E}\vec{Y}\right)\left(\vec{Y}-\mathbb{E}\vec{Y}\right)^{T}=\mathbb{E}\left(B\left(\vec{X}-\mathbb{E}\vec{X}\right)\left(\vec{X}-\mathbb{E}\vec{X}\right)^{T}B^{T}\right)=$$$$B\mathbb{E}\left(\vec{X}-\mathbb{E}\vec{X}\right)\left(\vec{X}-\mathbb{E}\vec{X}\right)^{T}B^{T}=B\Sigma B^{T}$$