Gaussian distribution is isotropic? I was in a seminar today and the lecturer said that the gaussian distribution is isotropic. What does it mean for a distribution to be isotropic? It seems like he is using this property for the pseudo-independence of vectors where each entry is sampled from the normal distribution.
 A: I'd just like to add a bit of visuals to the other answers.
When the variables are independent, i.e. the distrubtion is isotropic, it means that the distribution is aligned with the axis.
For example, for $\Sigma = \begin{pmatrix}1 & 0 \\ 0 & 30\end{pmatrix}$, you'd get something like this:

So, what happens when it is not isotropic? For example, when $\Sigma = \begin{pmatrix}1 & 15 \\ 15 & 30\end{pmatrix}$, the distribution appears "rotated", no longer aligned with the axes:

Note that this is just an example, the $\Sigma$ above is invalid since it is not PSD.

Code:
import numpy as np
from matplotlib import pyplot as plt

pts = np.random.multivariate_normal([0, 0], [[1,15],[15,31]], size=10000, check_valid='warn')

plt.scatter(pts[:, 0], pts[:, 1], s=1)
plt.xlim((-30,30))
plt.ylim((-30,30))

A: TLDR: An isotropic gaussian is one where the covariance matrix is represented by the simplified matrix $\Sigma = \sigma^{2}I$.
Some motivations:
Consider the traditional gaussian distribution:
$$
\mathcal{N}(\mu,\,\Sigma)
$$
where $\mu$ is the mean and $\Sigma$ is the covariance matrix.
Consider how the number of free parameters in this Gaussian grows as the number of dimensions grows.
$\mu$ will have a linear growth.
$\Sigma$ will have a quadratic growth!
This quadratic growth can be very computationally expensive, so $\Sigma$ is often restricted as $\Sigma = \sigma^{2}I$ where $\sigma^{2}I$ is a scalar variance multiplied by an identity matrix.
Note that this results in $\Sigma$ where all dimensions are independent and where the variance of each dimension is the same. So the gaussian will be circular/spherical.
Disclaimer: Not a mathematician, and I only just learned about this so may be missing some things :)
Hope that helps!
A: I am not a math major student but I will give a try to describe my understanding: an isotropic gaussian distribution means a multidimensional gaussian distribution with its variance matrix as an identity matrix multiplied by the same number on its diagonal. Each dimension can be seen as an independent one-dimension gaussian distribution (no covariance exists). 
A: Thanks to Tomoiagă, valuable learning opportunity. Explanation for psd in his answer.
w.r.t material in course Probabilistic Deep Learning with TensorFlow 2 in coursera.
the definition for positive semi-definite:

A symmetric matrix $M \in \mathbb{R}^{d\times d}$ is positive
semi-definite if it satisfies $b^TMb \ge 0$ for all nonzero
$b\in\mathbb{R}^d$. If, in addition, we have $b^TMb = 0 \Rightarrow
> b=0$ then $M$ is positive definite.

In one word: the valid covariance matrix should be symmetry and positive (semi-)definite. However, how to check the $\Sigma$ satisfied with the requirement?
Here comes The Cholesky decomposition

For every real-valued symmetric positive-definite matrix $M$, there is a unique lower-diagonal matrix $L$ that has  positive diagonal entries for which
\begin{equation}
     LL^T = M
 \end{equation}
This is called the Cholesky decomposition of $M$

Let's build some codes.
Given a psd matrix $ \Sigma = \begin{bmatrix} 10 & 5 \\ 5 & 10 \end{bmatrix}$ and a non-psd $ \Sigma = \begin{bmatrix} 10 & 11 \\ 11 & 10 \end{bmatrix}$ as the covariance matrix
sigma = [[10., 5.],[5., 10.]]
np.linalg.cholesky(sigma)

Output:
<tf.Tensor: shape=(2, 2), dtype=float32, numpy= array([[3.1622777, 0.  ],
       [1.5811388, 2.738613 ]], dtype=float32)>

bad_sigma = [[10., 11.], [11., 10.]]
try:
    scale_tril = tf.linalg.cholesky(bad_sigma)
except Exception as e:
    print(e)

Output:
Cholesky decomposition was not successful. The input might not be valid.

For convenience, a lower-triangular matrix is easier to create.
Last, the demo with isotropic Gaussian and non-isotropic ones.
import seaborn as sns
import matplotlib.pyplot as plt
%matplotlib inline

## isotropic normal
sigma = [[1., 0.],[0., 1.]]
lower_triangular = np.linalg.cholesky(sigma)
print(lower_triangular)
sigma = np.matmul(lower_triangular, np.transpose(lower_triangular))
##

bivariate_normal = np.random.multivariate_normal([0, 0], sigma, size=10000, check_valid='warn')

x1 = bivariate_normal[:, 0]
x2 = bivariate_normal[:, 1]
sns.jointplot(x1, x2, kind='kde', space=0, color='b')

isotropic-normal
# #non-isotropic normal
# sigma = [[1. , 0.6], [0.6, 1.]]

non-isotropic-normal
