The linear transformation of a vector $v$ is given by a matrix-vector multiplication $Tv$, that is $T_{n\times n}$ is the linear map.
This can be expressed as
$$\begin{bmatrix}
|&|&\cdots&| \\
T_{1\downarrow} &T_{2\downarrow}&\cdots& T_{n\downarrow} \\
|&|&\cdots&| \\
\end{bmatrix}
\begin{bmatrix}
v_1\\v_2\\\vdots\\v_n
\end{bmatrix}
=v_1 T_{1\downarrow}+v_1 T_{2\downarrow}\cdots+v_1 T_{n\downarrow}
$$
A matrix AA is singular if and only if 00 is an eigenvalue.
Therefore if you got a (at least one) zero eigenvalue then your linear map is of lower rank, that is it is equivalent to a matrix with a (at least one) zero column, so $T:\mathbb{R}^n\to \mathbb{R}^m$ where $m<n$ and therefore it has no unique representation in $\mathbb{R}^n$ - a degree of freedom.
Where a matrix has a repeated eigenvalue, say $\lambda_1=\dots\lambda_k, k<n$. One says it is one of an algebraic multiplicity $k$. The sum of algebraic multiplicities should be equal $n$, but not necessarily the number of eigenvalues. An eigenvalue of an algebraic multiplicity $k$ may have $\le k$ eigenvectors . The number of eigenvectors of an eigenvalue is called geometric multiplicity of that eigenvalue. When the geometric multiplicity of an eigenvalue is equal to its algebraic multiplicity there is no problem. When the geometric multiplicity is less then algebraic multiplicity the matrix\transformation is not diagonalizable, that is the basis of eigenvectors is missing a vector(s). This\these missing vector(s) can be added - and called generalized eivenvector, but there is no unique way to do it - thus a degree of freedom and, of course, non uniqueness.
Note that due to orthogonality\unitarity of $UU^T=I$, $VV^T=I$ and
$(UV^T)(VU^T)=I$ which implies
$det(U)=det(U^T)=\pm1$, $det(V)=det(V^T)=\pm1$ and
$det(UV^T)=\pm1$ regardless of algebraic multiplicity of eigenvalues.
So if $det(UV^T)=-1$ we have $det(U)=-det(V)=\pm1$
Now the question is why they have different sign when there is a an eigenvalue that has a geometric multiplicity lower then its algebraic multiplicity. The answer is hidden in $det U \Sigma V^t$, see here.
