I have computed the singular value decomposition of the following matrix
$$A= \begin{bmatrix}1&2\\0&1\\-1&0\\\end{bmatrix}$$
here are the important findings below.
$$\Sigma=\left[\begin{matrix}1 & 0 \\ 0 & \sqrt 6 \\ 0 & 0\end{matrix}\right]$$
$$V=\left[\begin{matrix}-\frac{2}{\sqrt 5} & \frac{1}{\sqrt 5} \\ \frac{1}{\sqrt 5} & \frac{2}{\sqrt 5}\end{matrix}\right]$$
$$U=\left[\begin{matrix}0 & \sqrt{\frac 5 6} & \frac{1}{\sqrt 6} \\ \frac{1}{\sqrt 5} & \sqrt{\frac 2 {15}} & -\sqrt{\frac 2 3} \\ \frac{2}{\sqrt 5} & -\frac{1}{\sqrt 30} & \frac{1}{\sqrt 6}\end{matrix}\right]$$
Now here is the question I am trying to solve now,
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 A.
So how can I find the best rank-one approximation of the matrix A?
Best Answer
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.