Singular value decomposition of a 2×2 matrix

Compute $A^T A$.
$$A^T A = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 9 & 0 \\ 0 & 4 \end{pmatrix}.$$

Singular values. The eigenvalues are $9$ and $4$, so the singular values are $\sigma_1 = 3$ and $\sigma_2 = 2$.

SVD. $A = U\Sigma V^T$ with $U$ and $V$ orthogonal. Here $A$ is already diagonal, so we can take
$$U = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad \Sigma = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}, \qquad V = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Then $U\Sigma V^T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix} = A$.

Conclusion. The singular values are $3$ and $2$, and the SVD is given above.