Orthogonal vectors, $||lambda a+mu b||$












0












$begingroup$


If $a$ and $b$ are orthogonal, and $||a|| = ||b|| = 1$, calculate $||lambda a+ mu b||$.



$langle lambda a+ mu b, lambda a + mu b rangle = langle lambda a, lambda arangle + 2 langle lambda a, mu brangle + langle mu b, mu b rangle$



$langle lambda a, mu brangle =0$



Here is where I am unsure.



$lambda^2 langle a,a rangle + mu^2 langle b,b rangle = lambda^2+mu^2$



Is this correct?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 18:11










  • $begingroup$
    @Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
    $endgroup$
    – Lillys
    Dec 13 '18 at 18:13






  • 1




    $begingroup$
    Do you know the definition of inner product?
    $endgroup$
    – Will M.
    Dec 13 '18 at 18:57










  • $begingroup$
    That would be true iff the vectors were orthogonal.
    $endgroup$
    – Michael Hoppe
    Dec 13 '18 at 19:23










  • $begingroup$
    @WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
    $endgroup$
    – Lillys
    Dec 13 '18 at 20:19
















0












$begingroup$


If $a$ and $b$ are orthogonal, and $||a|| = ||b|| = 1$, calculate $||lambda a+ mu b||$.



$langle lambda a+ mu b, lambda a + mu b rangle = langle lambda a, lambda arangle + 2 langle lambda a, mu brangle + langle mu b, mu b rangle$



$langle lambda a, mu brangle =0$



Here is where I am unsure.



$lambda^2 langle a,a rangle + mu^2 langle b,b rangle = lambda^2+mu^2$



Is this correct?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 18:11










  • $begingroup$
    @Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
    $endgroup$
    – Lillys
    Dec 13 '18 at 18:13






  • 1




    $begingroup$
    Do you know the definition of inner product?
    $endgroup$
    – Will M.
    Dec 13 '18 at 18:57










  • $begingroup$
    That would be true iff the vectors were orthogonal.
    $endgroup$
    – Michael Hoppe
    Dec 13 '18 at 19:23










  • $begingroup$
    @WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
    $endgroup$
    – Lillys
    Dec 13 '18 at 20:19














0












0








0





$begingroup$


If $a$ and $b$ are orthogonal, and $||a|| = ||b|| = 1$, calculate $||lambda a+ mu b||$.



$langle lambda a+ mu b, lambda a + mu b rangle = langle lambda a, lambda arangle + 2 langle lambda a, mu brangle + langle mu b, mu b rangle$



$langle lambda a, mu brangle =0$



Here is where I am unsure.



$lambda^2 langle a,a rangle + mu^2 langle b,b rangle = lambda^2+mu^2$



Is this correct?






linear-algebra







share|cite|improve this question











$endgroup$




If $a$ and $b$ are orthogonal, and $||a|| = ||b|| = 1$, calculate $||lambda a+ mu b||$.



$langle lambda a+ mu b, lambda a + mu b rangle = langle lambda a, lambda arangle + 2 langle lambda a, mu brangle + langle mu b, mu b rangle$



$langle lambda a, mu brangle =0$



Here is where I am unsure.



$lambda^2 langle a,a rangle + mu^2 langle b,b rangle = lambda^2+mu^2$



Is this correct?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 18:46









Zach Langley

9731019




9731019










asked Dec 13 '18 at 17:58









LillysLillys

778




778












  • $begingroup$
    Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 18:11










  • $begingroup$
    @Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
    $endgroup$
    – Lillys
    Dec 13 '18 at 18:13






  • 1




    $begingroup$
    Do you know the definition of inner product?
    $endgroup$
    – Will M.
    Dec 13 '18 at 18:57










  • $begingroup$
    That would be true iff the vectors were orthogonal.
    $endgroup$
    – Michael Hoppe
    Dec 13 '18 at 19:23










  • $begingroup$
    @WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
    $endgroup$
    – Lillys
    Dec 13 '18 at 20:19


















  • $begingroup$
    Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 18:11










  • $begingroup$
    @Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
    $endgroup$
    – Lillys
    Dec 13 '18 at 18:13






  • 1




    $begingroup$
    Do you know the definition of inner product?
    $endgroup$
    – Will M.
    Dec 13 '18 at 18:57










  • $begingroup$
    That would be true iff the vectors were orthogonal.
    $endgroup$
    – Michael Hoppe
    Dec 13 '18 at 19:23










  • $begingroup$
    @WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
    $endgroup$
    – Lillys
    Dec 13 '18 at 20:19
















$begingroup$
Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
$endgroup$
– Decaf-Math
Dec 13 '18 at 18:11




$begingroup$
Can you please edit your question by writing the mathematical logic and steps in MathJax? The answer should rather be $sqrt{lambda^2 + mu^2}$, but in order for us to see where your mistake on this occurs, it's much more helpful if you provide us with the work you completed.
$endgroup$
– Decaf-Math
Dec 13 '18 at 18:11












$begingroup$
@Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
$endgroup$
– Lillys
Dec 13 '18 at 18:13




$begingroup$
@Decaf-Math yeah I will use math Jax I currently don’t have a full keyboard so many symbols are not possible, but I will correct later. Thanks forgot about the square root
$endgroup$
– Lillys
Dec 13 '18 at 18:13




1




1




$begingroup$
Do you know the definition of inner product?
$endgroup$
– Will M.
Dec 13 '18 at 18:57




$begingroup$
Do you know the definition of inner product?
$endgroup$
– Will M.
Dec 13 '18 at 18:57












$begingroup$
That would be true iff the vectors were orthogonal.
$endgroup$
– Michael Hoppe
Dec 13 '18 at 19:23




$begingroup$
That would be true iff the vectors were orthogonal.
$endgroup$
– Michael Hoppe
Dec 13 '18 at 19:23












$begingroup$
@WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
$endgroup$
– Lillys
Dec 13 '18 at 20:19




$begingroup$
@WillM. Function into r, or c, for which linearity ,positive defitness and symmetry are true. .
$endgroup$
– Lillys
Dec 13 '18 at 20:19










1 Answer
1






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oldest

votes


















1












$begingroup$

Given: $a$ and $b$ are orthogonal, so $langle a , b rangle = langle b , a rangle = 0$, and $| a | = | b | = 1$, which means $sqrt{langle a , a rangle }= sqrt{ langle b , b rangle }= 1$.



We want to evaluate:



begin{align*}
| lambda a + mu b |^2 &= langle lambda a + mu b , lambda a + mu b rangle \
&=langle lambda a , lambda a + mu b rangle + langle mu b , lambda a + mu b rangle \
&= langle lambda a , lambda a rangle + langle lambda a , mu b rangle + langle mu b , lambda a rangle + langle mu b , mu b rangle \
&= lambdaoverline{lambda} langle a , a rangle + lambdaoverline{mu}langle a , b rangle + muoverline{lambda}langle b, a rangle + muoverline{mu}langle b , b rangle \
&=lambdaoverline{lambda}(1) + lambdaoverline{mu}(0) + muoverline{lambda}(0) + muoverline{mu}(1) \
&=lambdaoverline{lambda} + muoverline{mu},
end{align*}
Hence,
$$| lambda a + mu b | = sqrt{lambdaoverline{lambda}+muoverline{mu}}$$
where $overline{xi}$ denotes the complex conjugate of $xi$, note if $xiinmathbb{R}$ then $overline{xi}=xi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
    $endgroup$
    – B. Mehta
    Dec 13 '18 at 21:18










  • $begingroup$
    @B.Mehta Yes, oops! Thanks for the catch:)
    $endgroup$
    – Dragonite
    Dec 13 '18 at 21:28











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Given: $a$ and $b$ are orthogonal, so $langle a , b rangle = langle b , a rangle = 0$, and $| a | = | b | = 1$, which means $sqrt{langle a , a rangle }= sqrt{ langle b , b rangle }= 1$.



We want to evaluate:



begin{align*}
| lambda a + mu b |^2 &= langle lambda a + mu b , lambda a + mu b rangle \
&=langle lambda a , lambda a + mu b rangle + langle mu b , lambda a + mu b rangle \
&= langle lambda a , lambda a rangle + langle lambda a , mu b rangle + langle mu b , lambda a rangle + langle mu b , mu b rangle \
&= lambdaoverline{lambda} langle a , a rangle + lambdaoverline{mu}langle a , b rangle + muoverline{lambda}langle b, a rangle + muoverline{mu}langle b , b rangle \
&=lambdaoverline{lambda}(1) + lambdaoverline{mu}(0) + muoverline{lambda}(0) + muoverline{mu}(1) \
&=lambdaoverline{lambda} + muoverline{mu},
end{align*}
Hence,
$$| lambda a + mu b | = sqrt{lambdaoverline{lambda}+muoverline{mu}}$$
where $overline{xi}$ denotes the complex conjugate of $xi$, note if $xiinmathbb{R}$ then $overline{xi}=xi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
    $endgroup$
    – B. Mehta
    Dec 13 '18 at 21:18










  • $begingroup$
    @B.Mehta Yes, oops! Thanks for the catch:)
    $endgroup$
    – Dragonite
    Dec 13 '18 at 21:28
















1












$begingroup$

Given: $a$ and $b$ are orthogonal, so $langle a , b rangle = langle b , a rangle = 0$, and $| a | = | b | = 1$, which means $sqrt{langle a , a rangle }= sqrt{ langle b , b rangle }= 1$.



We want to evaluate:



begin{align*}
| lambda a + mu b |^2 &= langle lambda a + mu b , lambda a + mu b rangle \
&=langle lambda a , lambda a + mu b rangle + langle mu b , lambda a + mu b rangle \
&= langle lambda a , lambda a rangle + langle lambda a , mu b rangle + langle mu b , lambda a rangle + langle mu b , mu b rangle \
&= lambdaoverline{lambda} langle a , a rangle + lambdaoverline{mu}langle a , b rangle + muoverline{lambda}langle b, a rangle + muoverline{mu}langle b , b rangle \
&=lambdaoverline{lambda}(1) + lambdaoverline{mu}(0) + muoverline{lambda}(0) + muoverline{mu}(1) \
&=lambdaoverline{lambda} + muoverline{mu},
end{align*}
Hence,
$$| lambda a + mu b | = sqrt{lambdaoverline{lambda}+muoverline{mu}}$$
where $overline{xi}$ denotes the complex conjugate of $xi$, note if $xiinmathbb{R}$ then $overline{xi}=xi$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
    $endgroup$
    – B. Mehta
    Dec 13 '18 at 21:18










  • $begingroup$
    @B.Mehta Yes, oops! Thanks for the catch:)
    $endgroup$
    – Dragonite
    Dec 13 '18 at 21:28














1












1








1





$begingroup$

Given: $a$ and $b$ are orthogonal, so $langle a , b rangle = langle b , a rangle = 0$, and $| a | = | b | = 1$, which means $sqrt{langle a , a rangle }= sqrt{ langle b , b rangle }= 1$.



We want to evaluate:



begin{align*}
| lambda a + mu b |^2 &= langle lambda a + mu b , lambda a + mu b rangle \
&=langle lambda a , lambda a + mu b rangle + langle mu b , lambda a + mu b rangle \
&= langle lambda a , lambda a rangle + langle lambda a , mu b rangle + langle mu b , lambda a rangle + langle mu b , mu b rangle \
&= lambdaoverline{lambda} langle a , a rangle + lambdaoverline{mu}langle a , b rangle + muoverline{lambda}langle b, a rangle + muoverline{mu}langle b , b rangle \
&=lambdaoverline{lambda}(1) + lambdaoverline{mu}(0) + muoverline{lambda}(0) + muoverline{mu}(1) \
&=lambdaoverline{lambda} + muoverline{mu},
end{align*}
Hence,
$$| lambda a + mu b | = sqrt{lambdaoverline{lambda}+muoverline{mu}}$$
where $overline{xi}$ denotes the complex conjugate of $xi$, note if $xiinmathbb{R}$ then $overline{xi}=xi$.






share|cite|improve this answer











$endgroup$



Given: $a$ and $b$ are orthogonal, so $langle a , b rangle = langle b , a rangle = 0$, and $| a | = | b | = 1$, which means $sqrt{langle a , a rangle }= sqrt{ langle b , b rangle }= 1$.



We want to evaluate:



begin{align*}
| lambda a + mu b |^2 &= langle lambda a + mu b , lambda a + mu b rangle \
&=langle lambda a , lambda a + mu b rangle + langle mu b , lambda a + mu b rangle \
&= langle lambda a , lambda a rangle + langle lambda a , mu b rangle + langle mu b , lambda a rangle + langle mu b , mu b rangle \
&= lambdaoverline{lambda} langle a , a rangle + lambdaoverline{mu}langle a , b rangle + muoverline{lambda}langle b, a rangle + muoverline{mu}langle b , b rangle \
&=lambdaoverline{lambda}(1) + lambdaoverline{mu}(0) + muoverline{lambda}(0) + muoverline{mu}(1) \
&=lambdaoverline{lambda} + muoverline{mu},
end{align*}
Hence,
$$| lambda a + mu b | = sqrt{lambdaoverline{lambda}+muoverline{mu}}$$
where $overline{xi}$ denotes the complex conjugate of $xi$, note if $xiinmathbb{R}$ then $overline{xi}=xi$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 21:27

























answered Dec 13 '18 at 21:07









DragoniteDragonite

1,058420




1,058420












  • $begingroup$
    It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
    $endgroup$
    – B. Mehta
    Dec 13 '18 at 21:18










  • $begingroup$
    @B.Mehta Yes, oops! Thanks for the catch:)
    $endgroup$
    – Dragonite
    Dec 13 '18 at 21:28


















  • $begingroup$
    It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
    $endgroup$
    – B. Mehta
    Dec 13 '18 at 21:18










  • $begingroup$
    @B.Mehta Yes, oops! Thanks for the catch:)
    $endgroup$
    – Dragonite
    Dec 13 '18 at 21:28
















$begingroup$
It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
$endgroup$
– B. Mehta
Dec 13 '18 at 21:18




$begingroup$
It looks like you're missing a square: you should have $|x|^2 = langle x, xrangle$.
$endgroup$
– B. Mehta
Dec 13 '18 at 21:18












$begingroup$
@B.Mehta Yes, oops! Thanks for the catch:)
$endgroup$
– Dragonite
Dec 13 '18 at 21:28




$begingroup$
@B.Mehta Yes, oops! Thanks for the catch:)
$endgroup$
– Dragonite
Dec 13 '18 at 21:28


















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