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Do the $|$ around $|langle u,vrangle|$ refer to absolute value in the inner product version of the Cauchy-Schwarz inequality?
Cauchy-Schwarz Inequality proof (for semi-inner-product A-module).Inequality involving inner product. $|langle u,vrangle+overlinelangle u,vrangle|le 2|langle u,vrangle|$Why $langle y,xrangle+langle x,yrangle=2mathrmRelangle x,yrangle$? And the rules of using absolute value, inner production and norm?Why is there an “absolute value” and a norm in the Schwarz Inequality?An inner product inequalityCauchy Schwarz inequality and absolute valueCauchy-Schwarz Inequality without using $langle a x,yrangle=alangle x,yrangle$Proof of Cauchy Schwarz InequalityQuestion regarding norms of Cauchy-Schwarz inequalityInequality involving inner product and norm
$begingroup$
The full inequality is:
$|langle u,vrangle| leq ||u|| ||v||$
I understand that $||$ around the vectors $u$ and $v$ signifies the taking of their norm, but what do the single | around $langle u,vrangle$ mean?
linear-algebra inner-product-space absolute-value convention
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
$endgroup$
add a comment |
$begingroup$
The full inequality is:
$|langle u,vrangle| leq ||u|| ||v||$
I understand that $||$ around the vectors $u$ and $v$ signifies the taking of their norm, but what do the single | around $langle u,vrangle$ mean?
linear-algebra inner-product-space absolute-value convention
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
$endgroup$
3
$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
1
$begingroup$
Note that in LaTeX, $|a|$ (| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.
$endgroup$
– Brian
Mar 29 at 16:51
add a comment |
$begingroup$
The full inequality is:
$|langle u,vrangle| leq ||u|| ||v||$
I understand that $||$ around the vectors $u$ and $v$ signifies the taking of their norm, but what do the single | around $langle u,vrangle$ mean?
linear-algebra inner-product-space absolute-value convention
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
$endgroup$
The full inequality is:
$|langle u,vrangle| leq ||u|| ||v||$
I understand that $||$ around the vectors $u$ and $v$ signifies the taking of their norm, but what do the single | around $langle u,vrangle$ mean?
linear-algebra inner-product-space absolute-value convention
linear-algebra inner-product-space absolute-value convention
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
asked Mar 29 at 16:43
James RonaldJames Ronald
3068
3068
3
$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
1
$begingroup$
Note that in LaTeX, $|a|$ (| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.
$endgroup$
– Brian
Mar 29 at 16:51
add a comment |
3
$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
1
$begingroup$
Note that in LaTeX, $|a|$ (| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.
$endgroup$
– Brian
Mar 29 at 16:51
3
3
$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
1
1
$begingroup$
Note that in LaTeX, $|a|$ (
| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.$endgroup$
– Brian
Mar 29 at 16:51
$begingroup$
Note that in LaTeX, $|a|$ (
| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.$endgroup$
– Brian
Mar 29 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, it is absolute value. Note that $langle u, v rangle$ is a scalar. In a real vector space, this is a real number, and you are taking its absolute value in the usual way. In a complex vector space, it's a complex number, and you are taking its complex modulus.
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
add a comment |
$begingroup$
Yes, they refer to the absolute value. Since $u$ and $v$ are elements of an inner product space, they require a specific norm. However, $langle u,v rangle$ is either a real or complex number, so we use the Euclidean norm.
answered Mar 29 at 16:58
SebSeb
114
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, it is absolute value. Note that $langle u, v rangle$ is a scalar. In a real vector space, this is a real number, and you are taking its absolute value in the usual way. In a complex vector space, it's a complex number, and you are taking its complex modulus.
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
add a comment |
$begingroup$
Yes, it is absolute value. Note that $langle u, v rangle$ is a scalar. In a real vector space, this is a real number, and you are taking its absolute value in the usual way. In a complex vector space, it's a complex number, and you are taking its complex modulus.
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
add a comment |
$begingroup$
Yes, it is absolute value. Note that $langle u, v rangle$ is a scalar. In a real vector space, this is a real number, and you are taking its absolute value in the usual way. In a complex vector space, it's a complex number, and you are taking its complex modulus.
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
Yes, it is absolute value. Note that $langle u, v rangle$ is a scalar. In a real vector space, this is a real number, and you are taking its absolute value in the usual way. In a complex vector space, it's a complex number, and you are taking its complex modulus.
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
answered Mar 29 at 16:54
Nate EldredgeNate Eldredge
64.5k682174
64.5k682174
add a comment |
add a comment |
$begingroup$
Yes, they refer to the absolute value. Since $u$ and $v$ are elements of an inner product space, they require a specific norm. However, $langle u,v rangle$ is either a real or complex number, so we use the Euclidean norm.
answered Mar 29 at 16:58
SebSeb
114
$endgroup$
add a comment |
$begingroup$
Yes, they refer to the absolute value. Since $u$ and $v$ are elements of an inner product space, they require a specific norm. However, $langle u,v rangle$ is either a real or complex number, so we use the Euclidean norm.
answered Mar 29 at 16:58
SebSeb
114
$endgroup$
add a comment |
$begingroup$
Yes, they refer to the absolute value. Since $u$ and $v$ are elements of an inner product space, they require a specific norm. However, $langle u,v rangle$ is either a real or complex number, so we use the Euclidean norm.
answered Mar 29 at 16:58
SebSeb
114
$endgroup$
Yes, they refer to the absolute value. Since $u$ and $v$ are elements of an inner product space, they require a specific norm. However, $langle u,v rangle$ is either a real or complex number, so we use the Euclidean norm.
answered Mar 29 at 16:58
SebSeb
114
answered Mar 29 at 16:58
SebSeb
114
answered Mar 29 at 16:58
SebSeb
114
answered Mar 29 at 16:58
SebSeb
114
114
add a comment |
add a comment |
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$begingroup$
Yes, it's absolute value.
$endgroup$
– avs
Mar 29 at 16:44
1
$begingroup$
Note that in LaTeX, $|a|$ (
| a |) should be used over $||a||$ (|| a ||) when typesetting vector norms. Compare the readability of $| v | |u | $ vs. $||u||||v||$.$endgroup$
– Brian
Mar 29 at 16:51