let $Y$ and $Z$ be two normal random independent variables. prove that: $Z$ and $\rho \frac{\sigma_X }{\sigma_Y } \cdot Y $ are also independent. let $ (X,Y) $ be a random Gaussian vector where:
\begin{align*}  X \sim \mathcal{N}(0,\sigma_X ^2 )  \end{align*}
\begin{align*}  Y \sim \mathcal{N}(0,\sigma_Y ^2 )  \end{align*}
and lets denote: $ Z = X - \rho \frac{\sigma_X }{\sigma_Y } Y $ where $ \rho $, the co-variance coefficient, is defined as:
\begin{align*} \rho = \frac{E(XY)}{\sigma_X\sigma_Y}  \end{align*} 
I had been asked to prove that $Y$ and $Z$ are independent random variables.
I proved it by finding matrix $A$ that satisfied the linear transofrmation:
$ \begin{pmatrix}Y\\ Z \\  \end{pmatrix} = 
\underbrace{\begin{pmatrix}0&1 \\ 1&- \rho \frac{\sigma_X }{\sigma_Y } \\  \end{pmatrix} }_{A}
 \begin{pmatrix}X\\ Y \\  \end{pmatrix} $ 
so that means $(Y,Z)$ is also a Gaussian vector.
by showing that $Cov(Y,Z)=0$ using:  $ Z = X - \rho \frac{\sigma_X }{\sigma_Y } Y $ and $\rho = \frac{E(XY)}{\sigma_X\sigma_Y}$, I was able to conclude that $Y$ and $Z$ are statistically independent 
variables.
Now, for the rest of the exercise that I'm doing, I must have also concluded from what I've done here that $Z$ is also independent of $ \hat X \overset{\Delta}{=}\rho \frac{\sigma_X }{\sigma_Y } \cdot Y $ 
this might be an easy question but I can't quiet grasp why this is really true, although intuitively it sounds true, but I couldn't prove this formally. 
all I could think of is theorems and rules that assume independence at the first place. If I want to do prove this, it must be by definition somehow, but is this the only way? and if yes, how can I start doing this? 
UPDATE: 
actually by thinking about this, I think I can do the same as I did before, saying:
$ \begin{pmatrix}Z\\ \hat X \\  \end{pmatrix} = 
\underbrace{\begin{pmatrix}1&0 \\ 0& \rho \frac{\sigma_X }{\sigma_Y } \\  \end{pmatrix} }_{A}
 \begin{pmatrix}Z\\ Y \\  \end{pmatrix} $
and showing that the $Cov(Z, \hat X) = 0$.
is this an accepted proof? 
 A: If you are aware of the following fact, it is easy:
If $X,Z$ are independent and $f:\mathbb{R}\to\mathbb{R}$ is a measurable function, then also $f(X),Z$ are independent.
(This follows from $\sigma(f(X),Z)\subset\sigma(X,Z)$ and the definition of independence for random variables via the generated sigma algebras.)
In your case $f:\mathbb{R}\to\mathbb{R}, x\mapsto \rho\frac{\sigma_X}{\sigma_Y}x$ is a linear function and hence measurable.
EDIT: Since you commented that you are not familiar with measure-theoretic probability theory, here is another way not using any measure theory.
Let X,Z be continuous independent random variables. And let $a\neq0$ be some constant real number. We want to show the $aX,Z$ is independent. The joint cdf is for all $(x,z)\in\mathbb{R}^2$
$$F_{aX,Z}(x,z)=F_{X,Z}(x/a,z)= F_{X}(x/a)F_{Z}(z)=F_{aX}(x)F_{Z}(z),$$
where the second inequality follows from the independence of $X,Z$. Hence, $X,aZ$ is independent.
In your example $a=\rho\frac{\sigma_X}{\sigma_Y}$.
A third way is to again argue that $(\hat{X},Z)$ is multivariate normal (same idea as what you have done for $(Y,Z)$) and show that the covariance is zero.
