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

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?

• The first column of $\mathbf{U}$ is determined by your matrix $\mathbf{A}$. The remaining columns of $\mathbf{U}$ can be any columns that are pairwise orthogonal and orthogonal to the first column. – megas Dec 8 '14 at 23:49
• The key here is to note that there are many possible choices for $U$. – Ben Grossmann Dec 8 '14 at 23:54
• yup, I guess so. However, when I use the maltab to check the answer, the value for U is always the same though. – cliff Dec 8 '14 at 23:59
• Because matlab is programmed to make a particular choice. Given a vector $u_1$, do you know how to find an orthonormal basis $\{u_1,u_2,u_3\}$? – Ben Grossmann Dec 9 '14 at 0:05
• Let me try the G-S method first, I will update my status – cliff Dec 9 '14 at 0:42

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 $$A$$.