Multiplication of a vector by an orthogonal matrix
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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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