I'm trying to draw an ellipsoid of the $3 \times 3$ covariance matrix. Usually, I see the sentence
an ellipsoid corresponding to the eigenvectors and eigenvalues of covariance matrix.
But from the equation
$$(\mathbf{x}-\boldsymbol\mu)^T\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu)\le \chi_{p,\alpha}^2 \tag{1}$$
in this answer, it seems to me that we do the eigen-decomposition of the inverse of covariance matrix.
Could you please elaborate on this point?
IMHO,
Let $\bar x$ and $\Sigma$ be the mean and covariance matrix of the data. Then the ellipsoid is $\{x \in \mathbb R^d | (x-\bar x)^T \Sigma(x-\bar x) = a\}$ for some $a>0$. The value of $a$ determines the confidence region. We consider the eigen-decomposition $\Sigma = P\Lambda P^{-1} =P\Lambda P^T$. Here $\Lambda$ is the diagonal matrix generated by eigenvalues $\lambda_1, \ldots, \lambda_d$, while each column of $P$ is an eigenvector. Let $y = P^T(x - \bar x)$. Then $(x-\bar x)^T \Sigma(x-\bar x) = y^T \Lambda y = \lambda_1y_1^2+...+\lambda_d y_d^2$. Then $y = P^T(x - \bar x)$ denotes the rotation of $x - \bar x$ by $P^T$.
