Multiplication of a vector by an orthogonal matrix












1












$begingroup$


I have a question, consider $V$ an orthogonal matrix, and $u$ and $z$ are vectors, and W is a matrix does :



$V'u = W V'z implies u = W z$ ?



I want to get rid of the orthogonal matrix $V'$, my intuition says that I can, but I don't know which property of the orthogonal matrices will help me to do say.
Thank you in advance.






linear-algebra matrices products







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think WZ is a scalar, while U is a vector. So they cannot be equal.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:09










  • $begingroup$
    no it's not a scalar, it's a vector
    $endgroup$
    – math geek
    Dec 6 '18 at 16:12
















1












$begingroup$


I have a question, consider $V$ an orthogonal matrix, and $u$ and $z$ are vectors, and W is a matrix does :



$V'u = W V'z implies u = W z$ ?



I want to get rid of the orthogonal matrix $V'$, my intuition says that I can, but I don't know which property of the orthogonal matrices will help me to do say.
Thank you in advance.






linear-algebra matrices products







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think WZ is a scalar, while U is a vector. So they cannot be equal.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:09










  • $begingroup$
    no it's not a scalar, it's a vector
    $endgroup$
    – math geek
    Dec 6 '18 at 16:12














1












1








1





$begingroup$


I have a question, consider $V$ an orthogonal matrix, and $u$ and $z$ are vectors, and W is a matrix does :



$V'u = W V'z implies u = W z$ ?



I want to get rid of the orthogonal matrix $V'$, my intuition says that I can, but I don't know which property of the orthogonal matrices will help me to do say.
Thank you in advance.






linear-algebra matrices products







share|cite|improve this question











$endgroup$




I have a question, consider $V$ an orthogonal matrix, and $u$ and $z$ are vectors, and W is a matrix does :



$V'u = W V'z implies u = W z$ ?



I want to get rid of the orthogonal matrix $V'$, my intuition says that I can, but I don't know which property of the orthogonal matrices will help me to do say.
Thank you in advance.






linear-algebra matrices products




linear-algebra matrices products






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 16:38









Roberto Rastapopoulos

896424




896424










asked Dec 6 '18 at 16:05









math geekmath geek

63




63








  • 1




    $begingroup$
    I think WZ is a scalar, while U is a vector. So they cannot be equal.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:09










  • $begingroup$
    no it's not a scalar, it's a vector
    $endgroup$
    – math geek
    Dec 6 '18 at 16:12














  • 1




    $begingroup$
    I think WZ is a scalar, while U is a vector. So they cannot be equal.
    $endgroup$
    – Andrei
    Dec 6 '18 at 16:09










  • $begingroup$
    no it's not a scalar, it's a vector
    $endgroup$
    – math geek
    Dec 6 '18 at 16:12








1




1




$begingroup$
I think WZ is a scalar, while U is a vector. So they cannot be equal.
$endgroup$
– Andrei
Dec 6 '18 at 16:09




$begingroup$
I think WZ is a scalar, while U is a vector. So they cannot be equal.
$endgroup$
– Andrei
Dec 6 '18 at 16:09












$begingroup$
no it's not a scalar, it's a vector
$endgroup$
– math geek
Dec 6 '18 at 16:12




$begingroup$
no it's not a scalar, it's a vector
$endgroup$
– math geek
Dec 6 '18 at 16:12










1 Answer
1






active

oldest

votes


















2












$begingroup$

Assuming $V'u = WV'z$, we have



$$V'(u - Wz) = (WV' - V'W) z.$$



For the left-hand side to be zero for arbitrary $z$, $W$ and $V'$ have to commute, so your statement is not true in general.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    this one V'(u - Wz) = (WV' - V'W) z.
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
    $endgroup$
    – Adrian Keister
    Dec 6 '18 at 16:50











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1 Answer
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1 Answer
1






active

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active

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active

oldest

votes









2












$begingroup$

Assuming $V'u = WV'z$, we have



$$V'(u - Wz) = (WV' - V'W) z.$$



For the left-hand side to be zero for arbitrary $z$, $W$ and $V'$ have to commute, so your statement is not true in general.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    this one V'(u - Wz) = (WV' - V'W) z.
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
    $endgroup$
    – Adrian Keister
    Dec 6 '18 at 16:50
















2












$begingroup$

Assuming $V'u = WV'z$, we have



$$V'(u - Wz) = (WV' - V'W) z.$$



For the left-hand side to be zero for arbitrary $z$, $W$ and $V'$ have to commute, so your statement is not true in general.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    this one V'(u - Wz) = (WV' - V'W) z.
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
    $endgroup$
    – Adrian Keister
    Dec 6 '18 at 16:50














2












2








2





$begingroup$

Assuming $V'u = WV'z$, we have



$$V'(u - Wz) = (WV' - V'W) z.$$



For the left-hand side to be zero for arbitrary $z$, $W$ and $V'$ have to commute, so your statement is not true in general.






share|cite|improve this answer









$endgroup$



Assuming $V'u = WV'z$, we have



$$V'(u - Wz) = (WV' - V'W) z.$$



For the left-hand side to be zero for arbitrary $z$, $W$ and $V'$ have to commute, so your statement is not true in general.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 16:24









Roberto RastapopoulosRoberto Rastapopoulos

896424




896424












  • $begingroup$
    yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    this one V'(u - Wz) = (WV' - V'W) z.
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
    $endgroup$
    – Adrian Keister
    Dec 6 '18 at 16:50


















  • $begingroup$
    yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    this one V'(u - Wz) = (WV' - V'W) z.
    $endgroup$
    – math geek
    Dec 6 '18 at 16:46










  • $begingroup$
    Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
    $endgroup$
    – Adrian Keister
    Dec 6 '18 at 16:50
















$begingroup$
yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
$endgroup$
– math geek
Dec 6 '18 at 16:46




$begingroup$
yes, my matrices do commute ! thank you for your answer, although i didn't fully understand why this equality is true :
$endgroup$
– math geek
Dec 6 '18 at 16:46












$begingroup$
this one V'(u - Wz) = (WV' - V'W) z.
$endgroup$
– math geek
Dec 6 '18 at 16:46




$begingroup$
this one V'(u - Wz) = (WV' - V'W) z.
$endgroup$
– math geek
Dec 6 '18 at 16:46












$begingroup$
Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
$endgroup$
– Adrian Keister
Dec 6 '18 at 16:50




$begingroup$
Another way of thinking about it: if $V'$ and $W$ commute, then $V'u=WV'z$ if and only if $V'u=V'Wz$ if and only if ${V'}^{dagger}V'u={V'}^{dagger}V'Wz$ if and only if $u=Wz.$
$endgroup$
– Adrian Keister
Dec 6 '18 at 16:50


















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