Mixing
Is one norm property: $p(v)=0 iff v=0$ or positive definiteness?
$begingroup$
I'm confused as to seeing different requirements for norms.
Triangle eq. and homogeneity seem to be in all, but the last property seems to be either:
$p(v)=0 iff v=0$ or positive definiteness
How are these related/equivalent?
It seems that positive definiteness may be able to capture the rule about norm having to be positive, while also capturing the zero element property, both at the same time?
norm
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
$endgroup$
add a comment |
$begingroup$
I'm confused as to seeing different requirements for norms.
Triangle eq. and homogeneity seem to be in all, but the last property seems to be either:
$p(v)=0 iff v=0$ or positive definiteness
How are these related/equivalent?
It seems that positive definiteness may be able to capture the rule about norm having to be positive, while also capturing the zero element property, both at the same time?
norm
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
$endgroup$
add a comment |
$begingroup$
I'm confused as to seeing different requirements for norms.
Triangle eq. and homogeneity seem to be in all, but the last property seems to be either:
$p(v)=0 iff v=0$ or positive definiteness
How are these related/equivalent?
It seems that positive definiteness may be able to capture the rule about norm having to be positive, while also capturing the zero element property, both at the same time?
norm
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
$endgroup$
I'm confused as to seeing different requirements for norms.
Triangle eq. and homogeneity seem to be in all, but the last property seems to be either:
$p(v)=0 iff v=0$ or positive definiteness
How are these related/equivalent?
It seems that positive definiteness may be able to capture the rule about norm having to be positive, while also capturing the zero element property, both at the same time?
norm
norm
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
asked Jan 19 at 12:45
mavaviljmavavilj
2,81411137
2,81411137
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Positive definiteness of $p$ states that $p(v) = 0 iff v = 0$ and $p(v) geq 0$ for all $v in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) geq 0$ for all $v in V$ is implicitly encoded since $p$ is a function $p: V to [0, infty)$, hence you only need to include $p(v) = 0 iff v = 0$ as an additional requirement.
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
$endgroup$
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
add a comment |
$begingroup$
Let $V$ be an $mathbb{R}$-vector space. A norm is a mapping $lVertcdotrVertcolon Vrightarrowmathbb{R}$, satisfying the following properties. For all $v,win V$, you have the triangle inequality $lVert v+wrVertlelVert vrVert+lVert wrVert$. For all $lambdainmathbb{R},vin V$ you have absolute homogenity, i.e. $lVert lambda vrVert=lvertlambdavertlVert vrVert$. Then we require either of the following properties:
$lVert vrVert=0Rightarrow v=0$ for $vin V$,
$lVert vrVertge0$ for all $vin V$ with equality iff $v=0$.
I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $lVert vrVert<0$ for some $vin V$, you would - by absolute homogenity - have $lVert-vrVert=lvert-1rvertlVert vrVert=lVert vrVert<0$ and then - by the triangle inequality - $0=lVert0rVert=lVert v+(-v)rVertlelVert vrVert+lVert-vrVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $lVert vrVert=0Leftrightarrow v=0$ then follows from the first property as $lVert0rVert=lVertlambda0rVert=lvertlambdarvertlVert0rVert$ for all $lambdainmathbb{R}$ and hence $lVert0rVert=0$.
answered Jan 19 at 13:05
ThorgottThorgott
600314
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Positive definiteness of $p$ states that $p(v) = 0 iff v = 0$ and $p(v) geq 0$ for all $v in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) geq 0$ for all $v in V$ is implicitly encoded since $p$ is a function $p: V to [0, infty)$, hence you only need to include $p(v) = 0 iff v = 0$ as an additional requirement.
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
$endgroup$
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
add a comment |
$begingroup$
Positive definiteness of $p$ states that $p(v) = 0 iff v = 0$ and $p(v) geq 0$ for all $v in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) geq 0$ for all $v in V$ is implicitly encoded since $p$ is a function $p: V to [0, infty)$, hence you only need to include $p(v) = 0 iff v = 0$ as an additional requirement.
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
$endgroup$
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
add a comment |
$begingroup$
Positive definiteness of $p$ states that $p(v) = 0 iff v = 0$ and $p(v) geq 0$ for all $v in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) geq 0$ for all $v in V$ is implicitly encoded since $p$ is a function $p: V to [0, infty)$, hence you only need to include $p(v) = 0 iff v = 0$ as an additional requirement.
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
$endgroup$
Positive definiteness of $p$ states that $p(v) = 0 iff v = 0$ and $p(v) geq 0$ for all $v in V$. This is definitely required for every norm.
If you're reading the Wikipedia article, $p(v) geq 0$ for all $v in V$ is implicitly encoded since $p$ is a function $p: V to [0, infty)$, hence you only need to include $p(v) = 0 iff v = 0$ as an additional requirement.
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
answered Jan 19 at 12:58
bitesizebobitesizebo
1,59618
1,59618
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
add a comment |
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
$begingroup$
Indeed. I don't like that hidden requirement in the wiki article. In most similar definitions there is already an implicit (semi-hidden) requirement that the the operation must be defined for all elements of the set. The norm definition on wiki has another implicit requirement that is somewhat unexpected.
$endgroup$
– I like Serena
Jan 19 at 13:05
add a comment |
$begingroup$
Let $V$ be an $mathbb{R}$-vector space. A norm is a mapping $lVertcdotrVertcolon Vrightarrowmathbb{R}$, satisfying the following properties. For all $v,win V$, you have the triangle inequality $lVert v+wrVertlelVert vrVert+lVert wrVert$. For all $lambdainmathbb{R},vin V$ you have absolute homogenity, i.e. $lVert lambda vrVert=lvertlambdavertlVert vrVert$. Then we require either of the following properties:
$lVert vrVert=0Rightarrow v=0$ for $vin V$,
$lVert vrVertge0$ for all $vin V$ with equality iff $v=0$.
I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $lVert vrVert<0$ for some $vin V$, you would - by absolute homogenity - have $lVert-vrVert=lvert-1rvertlVert vrVert=lVert vrVert<0$ and then - by the triangle inequality - $0=lVert0rVert=lVert v+(-v)rVertlelVert vrVert+lVert-vrVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $lVert vrVert=0Leftrightarrow v=0$ then follows from the first property as $lVert0rVert=lVertlambda0rVert=lvertlambdarvertlVert0rVert$ for all $lambdainmathbb{R}$ and hence $lVert0rVert=0$.
answered Jan 19 at 13:05
ThorgottThorgott
600314
$endgroup$
add a comment |
$begingroup$
Let $V$ be an $mathbb{R}$-vector space. A norm is a mapping $lVertcdotrVertcolon Vrightarrowmathbb{R}$, satisfying the following properties. For all $v,win V$, you have the triangle inequality $lVert v+wrVertlelVert vrVert+lVert wrVert$. For all $lambdainmathbb{R},vin V$ you have absolute homogenity, i.e. $lVert lambda vrVert=lvertlambdavertlVert vrVert$. Then we require either of the following properties:
$lVert vrVert=0Rightarrow v=0$ for $vin V$,
$lVert vrVertge0$ for all $vin V$ with equality iff $v=0$.
I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $lVert vrVert<0$ for some $vin V$, you would - by absolute homogenity - have $lVert-vrVert=lvert-1rvertlVert vrVert=lVert vrVert<0$ and then - by the triangle inequality - $0=lVert0rVert=lVert v+(-v)rVertlelVert vrVert+lVert-vrVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $lVert vrVert=0Leftrightarrow v=0$ then follows from the first property as $lVert0rVert=lVertlambda0rVert=lvertlambdarvertlVert0rVert$ for all $lambdainmathbb{R}$ and hence $lVert0rVert=0$.
answered Jan 19 at 13:05
ThorgottThorgott
600314
$endgroup$
add a comment |
$begingroup$
Let $V$ be an $mathbb{R}$-vector space. A norm is a mapping $lVertcdotrVertcolon Vrightarrowmathbb{R}$, satisfying the following properties. For all $v,win V$, you have the triangle inequality $lVert v+wrVertlelVert vrVert+lVert wrVert$. For all $lambdainmathbb{R},vin V$ you have absolute homogenity, i.e. $lVert lambda vrVert=lvertlambdavertlVert vrVert$. Then we require either of the following properties:
$lVert vrVert=0Rightarrow v=0$ for $vin V$,
$lVert vrVertge0$ for all $vin V$ with equality iff $v=0$.
I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $lVert vrVert<0$ for some $vin V$, you would - by absolute homogenity - have $lVert-vrVert=lvert-1rvertlVert vrVert=lVert vrVert<0$ and then - by the triangle inequality - $0=lVert0rVert=lVert v+(-v)rVertlelVert vrVert+lVert-vrVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $lVert vrVert=0Leftrightarrow v=0$ then follows from the first property as $lVert0rVert=lVertlambda0rVert=lvertlambdarvertlVert0rVert$ for all $lambdainmathbb{R}$ and hence $lVert0rVert=0$.
answered Jan 19 at 13:05
ThorgottThorgott
600314
$endgroup$
Let $V$ be an $mathbb{R}$-vector space. A norm is a mapping $lVertcdotrVertcolon Vrightarrowmathbb{R}$, satisfying the following properties. For all $v,win V$, you have the triangle inequality $lVert v+wrVertlelVert vrVert+lVert wrVert$. For all $lambdainmathbb{R},vin V$ you have absolute homogenity, i.e. $lVert lambda vrVert=lvertlambdavertlVert vrVert$. Then we require either of the following properties:
$lVert vrVert=0Rightarrow v=0$ for $vin V$,
$lVert vrVertge0$ for all $vin V$ with equality iff $v=0$.
I've seen either of these two properties be referred to as positive-definiteness, so I won't name them. The second property trivially implies the former property. For the other direction, notice that if $lVert vrVert<0$ for some $vin V$, you would - by absolute homogenity - have $lVert-vrVert=lvert-1rvertlVert vrVert=lVert vrVert<0$ and then - by the triangle inequality - $0=lVert0rVert=lVert v+(-v)rVertlelVert vrVert+lVert-vrVert<0$, a contradiction, so the norm is definitely non-negative. The equivalency $lVert vrVert=0Leftrightarrow v=0$ then follows from the first property as $lVert0rVert=lVertlambda0rVert=lvertlambdarvertlVert0rVert$ for all $lambdainmathbb{R}$ and hence $lVert0rVert=0$.
answered Jan 19 at 13:05
ThorgottThorgott
600314
answered Jan 19 at 13:05
ThorgottThorgott
600314
answered Jan 19 at 13:05
ThorgottThorgott
600314
answered Jan 19 at 13:05
ThorgottThorgott
600314
600314
add a comment |
add a comment |
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