Proving Equality of the Induced Matrix Norm
I need to prove that the induced matrix norm satisfies $$|A| = max_{|x| = 1} |Ax|$$
Here's what I've done so far, and I'm not sure how to make the connection.
By definition, $$|A| = max_{xneq 0}{|Ax| over |x|}.$$ Because $xneq0$, $|x| neq 0 implies |x| = alpha > 0$ for any nonzero $xin mathbb{R}^n$ by definition of a vector norm. So, $$begin{align}|x| = alpha &iff {1 over alpha}|x| = 1 \ &iffleft|{1 over alpha}xright| = 1 \ &iff |hat{x}| = 1. tag{Let ${1 over alpha}x = hat{x}$}end{align}$$ I need to show that $$max_{|x| = 1}|Ax| = max_{xneq0}{|Ax| over |x|}.$$ Well I've shown that $$x neq 0 implies |hat x| = 1.$$ So does it suffice to say that since $|hat x| = 1$, $$|A| = max_{x neq 0}{|Ax| over |x|} = max_{|hat x| = 1}{|Ahat x | over |hat x|} = max_{|hat x| = 1} {|Ahat x| over 1} = max_{|hat x| = 1} |Ahat x|?$$
Thanks so much!
linear-algebra proof-verification proof-writing vectors numerical-linear-algebra
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
add a comment |
I need to prove that the induced matrix norm satisfies $$|A| = max_{|x| = 1} |Ax|$$
Here's what I've done so far, and I'm not sure how to make the connection.
By definition, $$|A| = max_{xneq 0}{|Ax| over |x|}.$$ Because $xneq0$, $|x| neq 0 implies |x| = alpha > 0$ for any nonzero $xin mathbb{R}^n$ by definition of a vector norm. So, $$begin{align}|x| = alpha &iff {1 over alpha}|x| = 1 \ &iffleft|{1 over alpha}xright| = 1 \ &iff |hat{x}| = 1. tag{Let ${1 over alpha}x = hat{x}$}end{align}$$ I need to show that $$max_{|x| = 1}|Ax| = max_{xneq0}{|Ax| over |x|}.$$ Well I've shown that $$x neq 0 implies |hat x| = 1.$$ So does it suffice to say that since $|hat x| = 1$, $$|A| = max_{x neq 0}{|Ax| over |x|} = max_{|hat x| = 1}{|Ahat x | over |hat x|} = max_{|hat x| = 1} {|Ahat x| over 1} = max_{|hat x| = 1} |Ahat x|?$$
Thanks so much!
linear-algebra proof-verification proof-writing vectors numerical-linear-algebra
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30
add a comment |
I need to prove that the induced matrix norm satisfies $$|A| = max_{|x| = 1} |Ax|$$
Here's what I've done so far, and I'm not sure how to make the connection.
By definition, $$|A| = max_{xneq 0}{|Ax| over |x|}.$$ Because $xneq0$, $|x| neq 0 implies |x| = alpha > 0$ for any nonzero $xin mathbb{R}^n$ by definition of a vector norm. So, $$begin{align}|x| = alpha &iff {1 over alpha}|x| = 1 \ &iffleft|{1 over alpha}xright| = 1 \ &iff |hat{x}| = 1. tag{Let ${1 over alpha}x = hat{x}$}end{align}$$ I need to show that $$max_{|x| = 1}|Ax| = max_{xneq0}{|Ax| over |x|}.$$ Well I've shown that $$x neq 0 implies |hat x| = 1.$$ So does it suffice to say that since $|hat x| = 1$, $$|A| = max_{x neq 0}{|Ax| over |x|} = max_{|hat x| = 1}{|Ahat x | over |hat x|} = max_{|hat x| = 1} {|Ahat x| over 1} = max_{|hat x| = 1} |Ahat x|?$$
Thanks so much!
linear-algebra proof-verification proof-writing vectors numerical-linear-algebra
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
I need to prove that the induced matrix norm satisfies $$|A| = max_{|x| = 1} |Ax|$$
Here's what I've done so far, and I'm not sure how to make the connection.
By definition, $$|A| = max_{xneq 0}{|Ax| over |x|}.$$ Because $xneq0$, $|x| neq 0 implies |x| = alpha > 0$ for any nonzero $xin mathbb{R}^n$ by definition of a vector norm. So, $$begin{align}|x| = alpha &iff {1 over alpha}|x| = 1 \ &iffleft|{1 over alpha}xright| = 1 \ &iff |hat{x}| = 1. tag{Let ${1 over alpha}x = hat{x}$}end{align}$$ I need to show that $$max_{|x| = 1}|Ax| = max_{xneq0}{|Ax| over |x|}.$$ Well I've shown that $$x neq 0 implies |hat x| = 1.$$ So does it suffice to say that since $|hat x| = 1$, $$|A| = max_{x neq 0}{|Ax| over |x|} = max_{|hat x| = 1}{|Ahat x | over |hat x|} = max_{|hat x| = 1} {|Ahat x| over 1} = max_{|hat x| = 1} |Ahat x|?$$
Thanks so much!
linear-algebra proof-verification proof-writing vectors numerical-linear-algebra
linear-algebra proof-verification proof-writing vectors numerical-linear-algebra
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
edited Apr 3 '16 at 8:17
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
asked Apr 3 '16 at 8:09
Decaf-Math
3,179825
3,179825
Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30
add a comment |
Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30
Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30
add a comment |
1 Answer
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Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have
$$
max_{|x| = 1} |Ax| = max_{|x| = 1} frac{|Ax|}{|x|} le max_{x not= 0} frac{|Ax|}{|x|}.
$$
For the reverse inequality, fix $x not= 0$. Then
$$
frac{|Ax|}{|x|} = left|Afrac{x}{|x|}right| le max_{|y| = 1} |Ay|,
$$
and so taking the maximum over nonzero $x$ gives the result.
answered Apr 3 '16 at 8:37
Aidan Sims
84959
add a comment |
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1 Answer
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1 Answer
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active
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votes
Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have
$$
max_{|x| = 1} |Ax| = max_{|x| = 1} frac{|Ax|}{|x|} le max_{x not= 0} frac{|Ax|}{|x|}.
$$
For the reverse inequality, fix $x not= 0$. Then
$$
frac{|Ax|}{|x|} = left|Afrac{x}{|x|}right| le max_{|y| = 1} |Ay|,
$$
and so taking the maximum over nonzero $x$ gives the result.
answered Apr 3 '16 at 8:37
Aidan Sims
84959
add a comment |
Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have
$$
max_{|x| = 1} |Ax| = max_{|x| = 1} frac{|Ax|}{|x|} le max_{x not= 0} frac{|Ax|}{|x|}.
$$
For the reverse inequality, fix $x not= 0$. Then
$$
frac{|Ax|}{|x|} = left|Afrac{x}{|x|}right| le max_{|y| = 1} |Ay|,
$$
and so taking the maximum over nonzero $x$ gives the result.
answered Apr 3 '16 at 8:37
Aidan Sims
84959
add a comment |
Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have
$$
max_{|x| = 1} |Ax| = max_{|x| = 1} frac{|Ax|}{|x|} le max_{x not= 0} frac{|Ax|}{|x|}.
$$
For the reverse inequality, fix $x not= 0$. Then
$$
frac{|Ax|}{|x|} = left|Afrac{x}{|x|}right| le max_{|y| = 1} |Ay|,
$$
and so taking the maximum over nonzero $x$ gives the result.
answered Apr 3 '16 at 8:37
Aidan Sims
84959
Just to add to the previous answer: a clearer argument (I think) is to proceed in two steps. You certainly have
$$
max_{|x| = 1} |Ax| = max_{|x| = 1} frac{|Ax|}{|x|} le max_{x not= 0} frac{|Ax|}{|x|}.
$$
For the reverse inequality, fix $x not= 0$. Then
$$
frac{|Ax|}{|x|} = left|Afrac{x}{|x|}right| le max_{|y| = 1} |Ay|,
$$
and so taking the maximum over nonzero $x$ gives the result.
answered Apr 3 '16 at 8:37
Aidan Sims
84959
answered Apr 3 '16 at 8:37
Aidan Sims
84959
answered Apr 3 '16 at 8:37
Aidan Sims
84959
answered Apr 3 '16 at 8:37
Aidan Sims
84959
84959
add a comment |
add a comment |
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Looks good for me. Neglecting the typo $||x||=a$ which should be $||x||=alpha$.
– robert.marik.cz
Apr 3 '16 at 8:16
Whoops! Fixed that. Thanks!
– Decaf-Math
Apr 3 '16 at 8:17
Hint.${x:|x|=1}={y/|y|:yne 0}$ and ${y:yne 0}={r x:|x|=1land 0ne rin R}.$..... Remark:We usually write $sup$ rather than $ max$ in the def'n of $|A|$ because in infinite-dimensional spaces the $max$ is not always attained.
– DanielWainfleet
Apr 3 '16 at 8:22
Ah, I'd forgotten that a magnitude of 1 meant that it was a unit vector. And I've used the $sup$ in norms before in my previous numerical analysis class with the $infty$ norm. It's just our textbook uses the $max$ definition.
– Decaf-Math
Apr 3 '16 at 8:30