Adding covariance to a multi-dimension random vector equivalence For a 1D random variable X, we know that $\mathrm{Var}(cX) = c^2 \mathrm{Var}(X)$ and $\mathrm{Var}(X+c) = \mathrm{Var}(X)$.  

I'm reading some notes that talk about a model defining a joint distribution on $(x,z)$ as
$$z \sim \mathcal{N}(0,I) \\
x|z \sim \mathcal{N}(\mu + \Lambda z, \Psi)$$  

where the parameters of my model are the vector $\mu \in\mathbb{R}^n$, the matrix $\Lambda \in \mathbb{R^{n\times k}}$ and the diagonal matrix $\Psi \in \mathbb{R}^{n\times n}$.  

They say that this is equivalent to the model
$$z \sim \mathcal{N}(0,I)\\ \epsilon \sim \mathcal{N}(0,\Psi)\\ x = \mu + \Lambda z + \epsilon
$$

where $\epsilon$ and $z$ are independent. They reason that this is true because we first simulate $z$, then simulate $x$ given $z$ and then ADD a covariance $\Psi$ noise to $\mu + \Lambda z$.  

I don't get how it's equivalent, and why we "add" the noise when adding constants to a variance keeps it unchanged.  

Can someone clarify how they're equivalent as well? with an intuition or possibly a mathematical proof?
If $\epsilon$ was not added to $x$, then $x$ could not have been a random variable given $z$, since $z$ would have determined $x$ un-randomly. To say why these forms are equivalent, note that if $$x=\mu+\Lambda z+\epsilon$$then $$x|z{\sim \mathcal{N}(\mu+\Lambda z+\mu_{\epsilon},\sigma^2_{\epsilon})\\\sim \mathcal{N}(\mu+\Lambda z,\Psi)}$$since given $z$ the quantity $\mu+\Lambda z$ serves as a deterministic number. In contrast if $$x=\mu+\Lambda z$$then $$x|z\sim\mathcal{N}(\mu+\Lambda z,0)$$which is a zero variance random variable or technically, a fixed scalar.

Alternative Intuition

Another intuition justifying why there should be two sources of randomness ($z$ and $\epsilon$) is that the random variable $x|z$ passes two steps; first: $z$ is randomly determined (first source of randomness); second: after determining $z$, the PDF of $x$ can be determined from $\mathcal{N}(\mu+\Lambda z,\Psi)$. Now $x$ will be determined obeying this PDF randomly (second source of randomness). Any equivalent equation for $x$ then, must have 2 sources of randomness, intuitively.
To show that both are equivalent - we first show that they both follow a multi-variate normal distribution. Then we show they have the same mean and covariance matrix. 

It is trivial to see that both the distributions follow a normal distribution. 

For the first representation, lets use the expectation and variance decomposition formulas:

The mean of a random variable $X$ is 
$$\Bbb E(X) = \Bbb E_z(\Bbb E(X| Z=z)) = \Bbb{E}_z(\mu+\Lambda z) = \mu$$

The covariance of the random variable $X$ is 
\begin{align}
\text{var}(X) &amp; = \Bbb E(\text{var}(X|Z)) + \text{var}(E(X|Z)) \\ &amp; = \Bbb{E}(\Psi) + \text{var}(\mu + \Lambda z) \\ &amp; = \Psi + \Lambda \Lambda^T
\end{align} 

For the second representation, 
the mean is $\mu$ because $z$ and $\epsilon$ are independent and expectation is a linear operator. The covariance for $X$ is $\Psi + \Lambda \Lambda^T$ because the variance of two independent variables is sum of their individual variances - here the indepedent variables are $\Lambda z$ and $\epsilon$.

Hence they are equivalent.