How $frac{c:e^{jtheta}- a^H x}{|a|_2^2}a rightarrow$ $frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$ for $theta$...
$begingroup$
I am reading a paper, there it rewrites this equation
$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$
into
$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$
Sorry, I just don't understand how to derive the latter from former. Please help.
My question in other words, how
$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?
linear-algebra
asked Dec 3 '18 at 10:19
learninglearning
566
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add a comment |
$begingroup$
I am reading a paper, there it rewrites this equation
$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$
into
$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$
Sorry, I just don't understand how to derive the latter from former. Please help.
My question in other words, how
$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?
linear-algebra
asked Dec 3 '18 at 10:19
learninglearning
566
$endgroup$
$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57
add a comment |
$begingroup$
I am reading a paper, there it rewrites this equation
$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$
into
$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$
Sorry, I just don't understand how to derive the latter from former. Please help.
My question in other words, how
$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?
linear-algebra
asked Dec 3 '18 at 10:19
learninglearning
566
$endgroup$
I am reading a paper, there it rewrites this equation
$$frac{c:e^{jtheta}- a^H x}{|a|_2^2}a$$
into
$$frac{c- |a^H x|}{|a|_2^2 : |a^H x|}aa^H x$$ if $theta$ to be angle of $a^H x$, where
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$
Sorry, I just don't understand how to derive the latter from former. Please help.
My question in other words, how
$e^{jtheta} = frac{a^H x}{|a^H x|}$ ?
linear-algebra
linear-algebra
asked Dec 3 '18 at 10:19
learninglearning
566
asked Dec 3 '18 at 10:19
learninglearning
566
edited Dec 3 '18 at 10:57
learning
asked Dec 3 '18 at 10:19
learninglearning
566
asked Dec 3 '18 at 10:19
learninglearning
566
asked Dec 3 '18 at 10:19
learninglearning
566
566
$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57
add a comment |
$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57
$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:
A picture is worth a thousand words:
The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
$endgroup$
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:
A picture is worth a thousand words:
The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
$endgroup$
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
add a comment |
$begingroup$
It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:
A picture is worth a thousand words:
The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
$endgroup$
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
add a comment |
$begingroup$
It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:
A picture is worth a thousand words:
The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
$endgroup$
It is not entirely clear from your question what $j$, $c$, $a$, $x$ and $a^H$ are, but I hope this helps:
A picture is worth a thousand words:
The point marked on the circle in the complex plane is $e^{itheta}$, where $theta$ is the angle that $z$ makes with the positive real axis. But it is also $frac{z}{|z|}$, because this is the vector $z$ scaled to lie on the unit circle. After all
$$left|frac{z}{|z|}right|=frac{|z|}{|z|}=1.$$
This shows that $e^{itheta}=frac{z}{|z|}$.
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
edited Dec 3 '18 at 10:54
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
answered Dec 3 '18 at 10:43
ServaesServaes
25.7k33996
25.7k33996
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
add a comment |
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
$begingroup$
This explains, thank you so much
$endgroup$
– learning
Dec 3 '18 at 10:58
add a comment |
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$begingroup$
What is $j$? What is $c$? What is $a$? What is $x$? What doe $a^H$ mean? Are these complex numbers, complex vectors, something else?
$endgroup$
– Servaes
Dec 3 '18 at 10:50
$begingroup$
$j = sqrt{-1}$, $c in mathbb{R}$, and $a, x in mathbb{C}^n$. Also, $a^H$ corresponds to complex conjugate transpose
$endgroup$
– learning
Dec 3 '18 at 10:57