The less than full rank case, can be done with SVD decomposition?

Question

The matrix is $$ A = \left( \begin{matrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{matrix} \right), $$ The rank is 1, there only one nonzero eigenvalue, and when I was doing the svd decomposition, I can only find the V and but not U. In U, I can only get the first column, but the rest two column is hard to calculate. As a look up online, the case is mostly column full rank or row full rank, or the Eigenvector is unit vector, but this one is quite special. anyone know how can I get the U here?

Answer 1

I'm going to use a shortcut to get the singular value decomposition here; it turns out that it's really easy for rank $1$ matrices. In particular, we have $$ A = \pmatrix{1\\2\\3} \pmatrix{1&2} = \sqrt{70} \pmatrix{1/\sqrt{14}\\2/\sqrt{14}\\3/\sqrt{14}} \pmatrix{1/\sqrt{5}&2/\sqrt{5}} $$ That doesn't look quite like an SVD... yet. Let $u_1$ and $v_1$ be the vectors given by $$ u_1 = \pmatrix{1/\sqrt{14}\\2/\sqrt{14}\\3/\sqrt{14}}, \quad v_1 = \pmatrix{1/\sqrt{5}\\2/\sqrt{5}} $$ We can then write $A = \sigma_1 u_1 v_1^T$, where $\sigma_1 = \sqrt{70}$.

Now, find a $u_2,u_3$ and $v_2$ so that $\{u_1,u_2,u_3\}$ and $\{v_1,v_2\}$ are bases of $\Bbb R^3$ and $\Bbb R^2$ respectively. Let $U$ be the matrix whose columns are $u_i$ and $V$ the matrix whose columns are $v_i$. Finally, let $$ \Sigma = \pmatrix{\sigma_1&0\\0&\sigma_2\\0&0} $$ where $\sigma_2 = 0$. We have $$ U\Sigma V^T = \pmatrix{u_1 & u_2 & u_3} \pmatrix{\sigma_1&0\\0&\sigma_2\\0&0} \pmatrix{v_1^T\\ v_2^T} $$ Compute this product using block-matrix multiplication and verify that I have given the SVD of $A$.