Is ${(a,b,c) in mathbb{C}^3 | a^3=b^3 }=S$ a subspace of $mathbb{C}^3$
$begingroup$
This is an exercise in Axler's "Linear Algebra Done Right" textbook, and the answer is no, because taking $x=(1,frac{-1+sqrt{3}i}{2},0)$ and $y=(1,frac{-1-sqrt{3}i}{2},0)$ we have $x+y notin S$.
Before looking up the answer, I was trying, and failing, to come up with a counterexample. Can someone explain the intuition behind this one? It looks like it was derived somehow.
linear-algebra
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
$endgroup$
add a comment |
$begingroup$
This is an exercise in Axler's "Linear Algebra Done Right" textbook, and the answer is no, because taking $x=(1,frac{-1+sqrt{3}i}{2},0)$ and $y=(1,frac{-1-sqrt{3}i}{2},0)$ we have $x+y notin S$.
Before looking up the answer, I was trying, and failing, to come up with a counterexample. Can someone explain the intuition behind this one? It looks like it was derived somehow.
linear-algebra
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
$endgroup$
4
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22
add a comment |
$begingroup$
This is an exercise in Axler's "Linear Algebra Done Right" textbook, and the answer is no, because taking $x=(1,frac{-1+sqrt{3}i}{2},0)$ and $y=(1,frac{-1-sqrt{3}i}{2},0)$ we have $x+y notin S$.
Before looking up the answer, I was trying, and failing, to come up with a counterexample. Can someone explain the intuition behind this one? It looks like it was derived somehow.
linear-algebra
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
$endgroup$
This is an exercise in Axler's "Linear Algebra Done Right" textbook, and the answer is no, because taking $x=(1,frac{-1+sqrt{3}i}{2},0)$ and $y=(1,frac{-1-sqrt{3}i}{2},0)$ we have $x+y notin S$.
Before looking up the answer, I was trying, and failing, to come up with a counterexample. Can someone explain the intuition behind this one? It looks like it was derived somehow.
linear-algebra
linear-algebra
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
asked Dec 23 '18 at 6:15
alwaysiamcaesaralwaysiamcaesar
525
525
4
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22
add a comment |
4
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22
4
4
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $a^3=b^3$, $c^3=d^3$ you want to know if
$$(a+c)^3 =(b+d)^3$$
This is not necessarily the case.
Taking cube-roots in $mathbb C$ is a multivalued operation.
$$x^3 = y^3 iff x=ysqrt[3]{1}$$
where the cube root is a cube root of one. There are three of them.
For intuition, consider the same idea over $mathbb R^3$ with $a^2=b^2$. Clearly this would not be a subspace because $(1,-1,0)$ and $(1,1,0)$ would belong to this set but their sum, $(2,0,0)$ would not.
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
add a comment |
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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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votes
$begingroup$
If $a^3=b^3$, $c^3=d^3$ you want to know if
$$(a+c)^3 =(b+d)^3$$
This is not necessarily the case.
Taking cube-roots in $mathbb C$ is a multivalued operation.
$$x^3 = y^3 iff x=ysqrt[3]{1}$$
where the cube root is a cube root of one. There are three of them.
For intuition, consider the same idea over $mathbb R^3$ with $a^2=b^2$. Clearly this would not be a subspace because $(1,-1,0)$ and $(1,1,0)$ would belong to this set but their sum, $(2,0,0)$ would not.
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
add a comment |
$begingroup$
If $a^3=b^3$, $c^3=d^3$ you want to know if
$$(a+c)^3 =(b+d)^3$$
This is not necessarily the case.
Taking cube-roots in $mathbb C$ is a multivalued operation.
$$x^3 = y^3 iff x=ysqrt[3]{1}$$
where the cube root is a cube root of one. There are three of them.
For intuition, consider the same idea over $mathbb R^3$ with $a^2=b^2$. Clearly this would not be a subspace because $(1,-1,0)$ and $(1,1,0)$ would belong to this set but their sum, $(2,0,0)$ would not.
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
add a comment |
$begingroup$
If $a^3=b^3$, $c^3=d^3$ you want to know if
$$(a+c)^3 =(b+d)^3$$
This is not necessarily the case.
Taking cube-roots in $mathbb C$ is a multivalued operation.
$$x^3 = y^3 iff x=ysqrt[3]{1}$$
where the cube root is a cube root of one. There are three of them.
For intuition, consider the same idea over $mathbb R^3$ with $a^2=b^2$. Clearly this would not be a subspace because $(1,-1,0)$ and $(1,1,0)$ would belong to this set but their sum, $(2,0,0)$ would not.
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
$endgroup$
If $a^3=b^3$, $c^3=d^3$ you want to know if
$$(a+c)^3 =(b+d)^3$$
This is not necessarily the case.
Taking cube-roots in $mathbb C$ is a multivalued operation.
$$x^3 = y^3 iff x=ysqrt[3]{1}$$
where the cube root is a cube root of one. There are three of them.
For intuition, consider the same idea over $mathbb R^3$ with $a^2=b^2$. Clearly this would not be a subspace because $(1,-1,0)$ and $(1,1,0)$ would belong to this set but their sum, $(2,0,0)$ would not.
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
edited Dec 23 '18 at 7:55
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
answered Dec 23 '18 at 7:01
David PetersonDavid Peterson
8,87821935
8,87821935
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
add a comment |
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Did you mean cube root of $+1$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 7:09
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
$begingroup$
Yes thank you, shubham
$endgroup$
– David Peterson
Dec 23 '18 at 7:55
add a comment |
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4
$begingroup$
$S$ is the union of three hyperplanes, given by $a=b$, $a=omega b$ and $a=omega^2b$ where $omega=exp(2pi i/3)$. If you take points from different hyperplanes, it's quite likely their sum is outside $S$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:22