4

I have computed the singular value decomposition (SVD) of the following matrix $A$.

$$ {\bf A} := \begin{bmatrix}1&2\\0&1\\-1&0\\\end{bmatrix} = \underbrace{\left[\begin{matrix}0 & \sqrt{\frac 5 6} & \frac{1}{\sqrt 6} \\ \frac{1}{\sqrt 5} & \sqrt{\frac 2 {15}} & -\sqrt{\frac 2 3} \\ \frac{2}{\sqrt 5} & -\frac{1}{\sqrt 30} & \frac{1}{\sqrt 6}\end{matrix}\right]}_{=: {\bf U}} \underbrace{\left[\begin{matrix}1 & 0 \\ 0 & \sqrt 6 \\ 0 & 0\end{matrix}\right]}_{=: {\bf \Sigma}} \underbrace{\left[\begin{matrix}-\frac{2}{\sqrt 5} & \frac{1}{\sqrt 5} \\ \frac{1}{\sqrt 5} & \frac{2}{\sqrt 5}\end{matrix}\right]^\top}_{=: {\bf V}^\top} $$

The singular value decomposition can be used to obtain the best rank $p\leq r $ approximation of a matrix $A$, by only keeping the first $p$ terms. Find the best rank-one approximation of the matrix $\bf A$. How can I find the best rank-one approximation of the matrix $\bf A$?

Rodrigo de Azevedo
  • 23,223
jh123
  • 1,438

1 Answers1

5

In $$ A=UΣV^T=\sum_{k=1}^d \sigma_ku_kv_k^T $$ you take, as you cited, only the term corresponding to the largest singular value. This is here the one for $k=2$, $\sigma_2=\sqrt6$, thus the approximation is $$ \sigma_2u_2v_2^T=\frac15\pmatrix{5\\2\\-1}\pmatrix{1&2} =\pmatrix{1&2\\\frac25&\frac45\\-\frac15&-\frac25}. $$ Note that in the usual convention the singular values are sorted in descending order so that the first singular value is the largest one and thus the first term of the SVD the best rank-1 approximation of the matrix.

Lutz Lehmann
  • 131,652
  • I am a little confused to what that means, what is the first term corresponding to the largest singular value then? – jh123 Oct 21 '18 at 17:26
  • okay so do I need to multiply those 3 to get the value? and what sign errors do I have in my SVD? – jh123 Oct 21 '18 at 17:36
  • Sorry, I automatically went with the usual convention that the first sing. value is the largest. Here the order is the other way, so the second is largest and thus you have to take the $k=2$ term. – Lutz Lehmann Oct 21 '18 at 17:40
  • does that mean my sign values are okay? – jh123 Oct 21 '18 at 17:40
  • I will accept your answer for your detailed explanation! – jh123 Oct 21 '18 at 17:41
  • Yes, the signs are ok, and now the approximation is really close to the original matrix except in position $(3,1)$, but on the other hand the quotient of the singular values is not that large, so that also the other term has something substantial to contribute. – Lutz Lehmann Oct 21 '18 at 17:42
  • great well thanks for everything! I appreciate your complete detailed solution, I will try and learn this method now! – jh123 Oct 21 '18 at 17:44
  • why is the second the largest instead of the first? – jh123 Oct 21 '18 at 17:50
  • 1
    Because $\sqrt6>1$. Note that the singular value is the spectral norm of the corresponding rank-1 term. Thus $|A-σ_2u_2v^T_2|=|σ_1u_1v^T_1|=σ_1=1$ is the smaller of the two variants. – Lutz Lehmann Oct 21 '18 at 17:54