How to calculate the SVD of a $3×3$ diagonal only matrix without computing the eigenvalues explicitly.

Question

Given $$A = \begin{bmatrix}1&0&0\\0&-2&0\\0&0&0\end{bmatrix} $$

What would be its SVD?

Key facts we can use: a) Singular values in the Sigma matrix are on the principal diagonal in decreasing order. b) Identity matrix is Unitary.

Answer 1

The eigenvalues of $AA^t$ are 1, 4 and 0. Therefore, the singular values of $A$ are equal to $2, 1, 0$, after sorting.

In such a simple case, the SVD can be obtained by rewriting $A$ as, where the $e_i$ are the natural basis vectors:

$$A = 1\, e_1 e_1^t -2\, e_2 e_2^t + 0\, e_3 e_3^t$$

and then the SVD in sum form:

$$ A = 2\, (-e_2) {e_2}^t + 1\, e_1 e_1^t + 0\, e_3 e_3^t$$

In equivalent matrix form:

$$A = \begin{bmatrix}0&1&0\\-1&0&0\\0&0&1\end{bmatrix} \begin{bmatrix}2&0&0\\0&1&0\\0&0&0\end{bmatrix} \begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix} $$