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What are $y^+$ and $y^-$ supposed to represent? $y$ is a vector.

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7  
Hard to say unless you give us some further context. –  lhf Mar 22 '11 at 1:06

1 Answer 1

up vote 6 down vote accepted

As lhf points out, it depends on what context you are using these.

One possible answer which I have seen is $y^{+}$ is a vector whose elements are same as $y$ if the corresponding element in $y$ is positive and is zero if the corresponding element in $y$ is negative. Similarly, $y^{-}$ is a vector whose elements are same as $y$ if the corresponding element in $y$ is negative and is zero if the corresponding element in $y$ is positive.i.e.

If $y \in \mathbb{R}^{n \times 1}$, then

$y^{+} \in \mathbb{R}^{n \times 1}$ with $y^{+}(i) = \max(y(i),0)$

and

$y^{-} \in \mathbb{R}^{n \times 1}$ with $y^{-}(i) = \min(y(i),0)$

Hence, $y(i) = y^{+}(i) + y^{-}(i)$

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Thank you, this is it. Sorry I didn't provide any context, but I'm too tired to type right now. –  Matt Gregory Mar 22 '11 at 1:18
    
FWIW, the book I'm using seems to be defining $y^-$ the way you did but negated so $y^-$ is nonnegative. –  Matt Gregory Mar 22 '11 at 1:24
1  
@Matt: Yes. Some people use that convention as well i.e. $y^{-}(i) = \max(-y(i),0)$ so that both $y^{+}$ and $y^{-}$ are non-negative and then we get $y(i) = y^{+}(i) - y^{-}(i)$ –  user17762 Mar 22 '11 at 6:12

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