First, the difference in the eigenvectors. Let $(\lambda,v)$ be an eigenpair of $A$, i.e., $A v = \lambda v$ and let $\alpha \in \mathbb{C} \setminus \{0\}$. Then
$$A (\alpha v) = \alpha A v = \alpha \lambda v = \lambda (\alpha v).$$
So, $v$ is an eigenvector of $A$ if and only if $\alpha v$ is an eigenvector of $A$. Both are equally "good", unless you desire some additional properties. Note that this works for any $A$, not just $A = C$.
Second, the significance of the left singular vectors is in computing the eigenvalue decomposition in $XX^T$ (in your notation: $X^T = X'$).
Third, a real diagonal matrix is orthogonal if and only if each of its diagonal elements is either $1$ or $-1$. Let us prove this.
Let $D = \mathop{\rm diag}(d_1,\dots,d_n)$. Obviously, $D = D^T$, so
$$D^TD = \mathop{\rm diag}(d_1^2,\dots,d_n^2).$$
So, $D^TD = {\rm I}$ if and only if $d_k^2 = 1$ for all $k$.
For complex matrices (and using complex adjoint $Q^*$ instead of transpose $Q^T$), we get that $|d_k| = 1$ for all $k$.
The singular vectors of a matrix $A$ are the eigenvectors of $A^* A$. In the case of a real symmetric matrix $B$, the eigenvectors of $B$ are eigenvectors of $B^* B = B^2$, but not vice versa (in the case where $\lambda$ and $-\lambda$ are both eigenvalues for some $\lambda \ne 0$).
To get the eigenvalues and eigenvectors of a matrix in Matlab, use eig.
Best Answer
Since you wrote in the comment to my other attempt at an answer that you want to visualize covariance matrices to see whether they are different, I'd recommend you simply plot the matrix entries coded by color. The plotting can be done by
pcolor, but for the comparison it is important to have a common colorscale.Assuming you have
nmmatrices of sizenvxnvstored as a three-dimensional arraycof dimensionnvxnvxnm:The result might look for example like this: