Find SVD of $A$

Question

How do I find the singular values? They somehow show that $\lambda_1 = 27, \lambda_2 = 6, \lambda_3 = 0$. I still can't see how they found them with the equations I made in my solution.

Answer 1

You have almost found the SVD of $A$

$$A = \hat{U} \Sigma \hat{V}^T = \begin{bmatrix} \frac{1}{\sqrt 6} & \frac{1}{\sqrt 3} & |\\ \frac{2}{\sqrt 6} & -\frac{1}{\sqrt 3} & \hat u_3\\ \frac{1}{\sqrt 6} & \frac{1}{\sqrt 3} & |\\\end{bmatrix} \begin{bmatrix} 3 \sqrt 3 & 0\\ 0 & \sqrt 6\\ 0 & 0\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}\\ \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\end{bmatrix}^T$$

The third column of $\hat U$ has not yet been determined. Since $\hat U$ must be an orthogonal matrix, its columns must be orthonormal. Hence, $\hat u_3$ must be orthogonal to the other two columns of $\hat U$

$$\begin{bmatrix} 1 & 2 & 1\\ 1 & -1 & 1\end{bmatrix} \begin{bmatrix} |\\ \hat u_3\\ |\end{bmatrix} = \begin{bmatrix} 0\\ 0\end{bmatrix}$$

Thus,

$$\hat u_3 = \begin{bmatrix} \frac{1}{\sqrt 2}\\ 0\\ -\frac{1}{\sqrt 2}\end{bmatrix}$$

Note that the eigendecomposition of $A A^T$ is

$$A A^T = \hat{U} \Sigma \hat{V}^T \hat{V} \Sigma^T \hat{U}^T = \hat{U} \Sigma \Sigma^T \hat{U}^T = \hat{U} \begin{bmatrix} 27 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 0\end{bmatrix} \hat{U}^T$$

Thus, the three eigenvalues of $A A^T$ are the squares of the two singular values of $A$ and zero.