norm of vectors u, and v transposed

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given vectors $u$ and $v$ in $R^n$ how do i prove that



$||uv^T||_2 = ||u||_2 ||v||_2$



is it sufficient to just calculate?



$||u||_2 = sqrtu^2_1+u^2_2 $



$||v||_2 = sqrtv^2_1+v^2_2 $



do i just multiply $u$ and $v^T$ and take the vector norm? or do i take the matrix norm here? or am i totally wrong and misunderstood the problem. Thanks in advance




linear-algebra numerical-linear-algebra




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  • There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
    – Saucy O'Path
    Aug 27 at 15:53






  • 1




    If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
    – David K
    Aug 27 at 15:53










  • "or do i take the matrix norm here?" Yes, on the left hand side.
    – amsmath
    Aug 27 at 15:57










  • @DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
    – kayaya10
    Aug 27 at 16:09














up vote
0
down vote

favorite












given vectors $u$ and $v$ in $R^n$ how do i prove that



$||uv^T||_2 = ||u||_2 ||v||_2$



is it sufficient to just calculate?



$||u||_2 = sqrtu^2_1+u^2_2 $



$||v||_2 = sqrtv^2_1+v^2_2 $



do i just multiply $u$ and $v^T$ and take the vector norm? or do i take the matrix norm here? or am i totally wrong and misunderstood the problem. Thanks in advance




linear-algebra numerical-linear-algebra




share|cite|improve this question




















  • There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
    – Saucy O'Path
    Aug 27 at 15:53






  • 1




    If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
    – David K
    Aug 27 at 15:53










  • "or do i take the matrix norm here?" Yes, on the left hand side.
    – amsmath
    Aug 27 at 15:57










  • @DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
    – kayaya10
    Aug 27 at 16:09












up vote
0
down vote

favorite









up vote
0
down vote

favorite











given vectors $u$ and $v$ in $R^n$ how do i prove that



$||uv^T||_2 = ||u||_2 ||v||_2$



is it sufficient to just calculate?



$||u||_2 = sqrtu^2_1+u^2_2 $



$||v||_2 = sqrtv^2_1+v^2_2 $



do i just multiply $u$ and $v^T$ and take the vector norm? or do i take the matrix norm here? or am i totally wrong and misunderstood the problem. Thanks in advance




linear-algebra numerical-linear-algebra




share|cite|improve this question












given vectors $u$ and $v$ in $R^n$ how do i prove that



$||uv^T||_2 = ||u||_2 ||v||_2$



is it sufficient to just calculate?



$||u||_2 = sqrtu^2_1+u^2_2 $



$||v||_2 = sqrtv^2_1+v^2_2 $



do i just multiply $u$ and $v^T$ and take the vector norm? or do i take the matrix norm here? or am i totally wrong and misunderstood the problem. Thanks in advance





linear-algebra numerical-linear-algebra





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 27 at 15:49









kayaya10

81




81











  • There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
    – Saucy O'Path
    Aug 27 at 15:53






  • 1




    If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
    – David K
    Aug 27 at 15:53










  • "or do i take the matrix norm here?" Yes, on the left hand side.
    – amsmath
    Aug 27 at 15:57










  • @DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
    – kayaya10
    Aug 27 at 16:09
















  • There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
    – Saucy O'Path
    Aug 27 at 15:53






  • 1




    If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
    – David K
    Aug 27 at 15:53










  • "or do i take the matrix norm here?" Yes, on the left hand side.
    – amsmath
    Aug 27 at 15:57










  • @DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
    – kayaya10
    Aug 27 at 16:09















There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
– Saucy O'Path
Aug 27 at 15:53




There is no theorem that says that for any two matrices $lVert ABrVert_2=lVert ArVert_2lVert BrVert_2$, because in general $lVert ABrVert_2gelVert ArVert_2lVert BrVert_2$ is false.
– Saucy O'Path
Aug 27 at 15:53




1




1




If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
– David K
Aug 27 at 15:53




If you "just multiply $u$ and $v^T$", what kind of object do you get as a result?
– David K
Aug 27 at 15:53












"or do i take the matrix norm here?" Yes, on the left hand side.
– amsmath
Aug 27 at 15:57




"or do i take the matrix norm here?" Yes, on the left hand side.
– amsmath
Aug 27 at 15:57












@DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
– kayaya10
Aug 27 at 16:09




@DavidK one get a matrix, which is why i wondered if we take the matrix norm. which seems to be the case since like amsmath confirmed
– kayaya10
Aug 27 at 16:09










1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










You may begin with the definition of the operator norm:
$$
|uv^T|_2=max_|uv^Tw|_2.
$$
Since $v^Tw$ is a scalar, we get
$$
|uv^T|_2=max_|v^Tw||u|_2
$$
and you may continue from here.



Edit. One may also begin with the equivalent definition that $|A|_2=sqrtrho(A^HA)$. In this case we have
$$
|uv^T|_2^2=rho(vu^Tuv^T)=|u|_2^2rho(vv^T)
$$
and it remains to prove that $rho(vv^T)=|v|_2^2$. This should be easy if you consider the images of $v$ and $v^perp$ under $vv^T$.






share|cite|improve this answer






















  • Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
    – kayaya10
    Aug 27 at 16:19











  • @kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
    – user1551
    Aug 27 at 17:07










  • Thank you so much this helped out!
    – kayaya10
    Aug 30 at 20:26










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










You may begin with the definition of the operator norm:
$$
|uv^T|_2=max_|uv^Tw|_2.
$$
Since $v^Tw$ is a scalar, we get
$$
|uv^T|_2=max_|v^Tw||u|_2
$$
and you may continue from here.



Edit. One may also begin with the equivalent definition that $|A|_2=sqrtrho(A^HA)$. In this case we have
$$
|uv^T|_2^2=rho(vu^Tuv^T)=|u|_2^2rho(vv^T)
$$
and it remains to prove that $rho(vv^T)=|v|_2^2$. This should be easy if you consider the images of $v$ and $v^perp$ under $vv^T$.






share|cite|improve this answer






















  • Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
    – kayaya10
    Aug 27 at 16:19











  • @kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
    – user1551
    Aug 27 at 17:07










  • Thank you so much this helped out!
    – kayaya10
    Aug 30 at 20:26














up vote
0
down vote



accepted










You may begin with the definition of the operator norm:
$$
|uv^T|_2=max_|uv^Tw|_2.
$$
Since $v^Tw$ is a scalar, we get
$$
|uv^T|_2=max_|v^Tw||u|_2
$$
and you may continue from here.



Edit. One may also begin with the equivalent definition that $|A|_2=sqrtrho(A^HA)$. In this case we have
$$
|uv^T|_2^2=rho(vu^Tuv^T)=|u|_2^2rho(vv^T)
$$
and it remains to prove that $rho(vv^T)=|v|_2^2$. This should be easy if you consider the images of $v$ and $v^perp$ under $vv^T$.






share|cite|improve this answer






















  • Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
    – kayaya10
    Aug 27 at 16:19











  • @kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
    – user1551
    Aug 27 at 17:07










  • Thank you so much this helped out!
    – kayaya10
    Aug 30 at 20:26












up vote
0
down vote



accepted







up vote
0
down vote



accepted






You may begin with the definition of the operator norm:
$$
|uv^T|_2=max_|uv^Tw|_2.
$$
Since $v^Tw$ is a scalar, we get
$$
|uv^T|_2=max_|v^Tw||u|_2
$$
and you may continue from here.



Edit. One may also begin with the equivalent definition that $|A|_2=sqrtrho(A^HA)$. In this case we have
$$
|uv^T|_2^2=rho(vu^Tuv^T)=|u|_2^2rho(vv^T)
$$
and it remains to prove that $rho(vv^T)=|v|_2^2$. This should be easy if you consider the images of $v$ and $v^perp$ under $vv^T$.






share|cite|improve this answer














You may begin with the definition of the operator norm:
$$
|uv^T|_2=max_|uv^Tw|_2.
$$
Since $v^Tw$ is a scalar, we get
$$
|uv^T|_2=max_|v^Tw||u|_2
$$
and you may continue from here.



Edit. One may also begin with the equivalent definition that $|A|_2=sqrtrho(A^HA)$. In this case we have
$$
|uv^T|_2^2=rho(vu^Tuv^T)=|u|_2^2rho(vv^T)
$$
and it remains to prove that $rho(vv^T)=|v|_2^2$. This should be easy if you consider the images of $v$ and $v^perp$ under $vv^T$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 27 at 17:14

























answered Aug 27 at 16:00









user1551

67.2k565123




67.2k565123











  • Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
    – kayaya10
    Aug 27 at 16:19











  • @kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
    – user1551
    Aug 27 at 17:07










  • Thank you so much this helped out!
    – kayaya10
    Aug 30 at 20:26
















  • Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
    – kayaya10
    Aug 27 at 16:19











  • @kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
    – user1551
    Aug 27 at 17:07










  • Thank you so much this helped out!
    – kayaya10
    Aug 30 at 20:26















Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
– kayaya10
Aug 27 at 16:19





Im sorry to say I dont quite understand what "operator" means. About the defintion of matrix norm in my textbook it says: $||A||_2 = sqrtp(A^H A $ where p is the spectral radius.
– kayaya10
Aug 27 at 16:19













@kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
– user1551
Aug 27 at 17:07




@kayaya10 That's the name of the norm. And the definition in your textbook is equivalent to the one here.
– user1551
Aug 27 at 17:07












Thank you so much this helped out!
– kayaya10
Aug 30 at 20:26




Thank you so much this helped out!
– kayaya10
Aug 30 at 20:26

















 

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