Problem: Suppose $A$ from $\mathbb{C}^3$ to $\mathbb{C}^3$ is given by $$ A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{array}\right]. $$
Question: How to find an orthonormal basis of $\mathbb{C}^3$ so that the matrix representation of $A$ consists of blocks of the form $$ \left[\begin{array}{cc} 0 & \lambda_i \\ \lambda_i & 0 \end{array}\right] $$ and otherwise all zeros?
Attempt so far: Suppose I try first to get a singular value decomposition $A$. Then $A^TA = V\Sigma^2 V^T$ and $AA^T = U\Sigma^2 U^T$. Since $A$ is skew-symmetric, we have that $A^TA = AA^T$ so that $U = V$ consisting of an orthonormal basis of eigenvectors for $AA^T$.
Now $AA^T$ has a eigenvalue of $0$ and a repeated eigenvalue of $14$ with two eigenvectors. I can use Gram-Schmidt to find an orthogonal basis from these eigenvectors and construct $V$.
But then I think I should be getting that $A = U\Sigma V^T = V\Sigma V^T$ but it's not working out that way. Here's $V$ and my $\Sigma$: $$ V=\left[\begin{array}{ccc} \dfrac{3}{\sqrt{14}} & -\dfrac{2}{\sqrt{13}} & -\dfrac{9}{\sqrt{1638}}\\ \dfrac{2}{\sqrt{14}}& \dfrac{3}{\sqrt{13}} & -\dfrac{6}{\sqrt{1638}} \\ \dfrac{1}{\sqrt{14}} & 0 & \dfrac{39}{\sqrt{1638}} \end{array}\right]. $$ $$ \Sigma=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & \sqrt{14} & 0 \\ 0 & 0 & \sqrt{14} \end{array}\right]. $$