Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In linear algebra, the single value decomposotion of $A_{mxn}$ with $rank(A) = r$ is defined as

$$ A_{mxn} = U\;\Sigma V = \sum^{r}_{k=1}u_k\sigma_kv_k^T $$

Is it true that for any $k=(1,\dots, r)$ the sum to the right in the above equation will always return the matrix $A_k$ with all zeros except for the row $k$ (where it has the values of A in the row k)? I have found this to be true for a number of examples but I am unsure if this holds as a general truth.

If you could include a brief explanation with your answer that would be great, thanks!

share|improve this question
    
Try $A = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$. Also, in the expression $U\Sigma V$, the matrix of singular values is usually denoted by the Greek capital letter sigma, $\Sigma$, not the summation symbol, $\sum$. –  Rahul Jan 15 '12 at 20:21
    
thx, fixed the sigma vs. sum issue. Thanks also for the counter example, guess I am wrong then. –  Robin Jan 15 '12 at 20:26

1 Answer 1

up vote 1 down vote accepted

It seems my comment has answered the question, so I'm posting it here with a little elaboration.

Let $A = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$. Its singular value decomposition is $A = U\Sigma V^T$, where $$\begin{align} U = V &= \frac1{\sqrt{2}}\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}, \\ \Sigma &= \begin{bmatrix}3 & 0 \\ 0 & 1\end{bmatrix}. \\ \end{align}$$ Assuming by $u_k$ and $v_k$ you mean the $k$th column of $U$ and $V$, we have $$\begin{align} u_1\sigma_1v_1^T &= \frac32\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}, \\ u_2\sigma_2v_2^T &= \frac12\begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}. \end{align}$$

In general, $u_k\sigma_kv_k^T$ is not a matrix with the $k$th row or column equal to that of $A$ and zeroes everywhere else. What it is, is a rank-one matrix that represents the following linear transformation: given an input vector $x$, take its component along $v_k$ and discard the rest, scale that by $\sigma_k$, and make it point along $u_k$ instead. (Apart from the scaling by $\sigma_k$, this is something like a rank-one version of an orthogonal transformation. I can't remember if there's a specific name for it.) What the singular value decomposition tells you is how to represent your rank-$r$ matrix as a linear combination of $r$ such rank-one linear transformations.

share|improve this answer
    
Thanks, that explains it very well! –  Robin Jan 16 '12 at 10:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.