Need help on understanding the solution (Linear Algebra done right)
up vote
2
down vote
favorite
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 at 8:18
JOHN
676
add a comment |
up vote
2
down vote
favorite
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 at 8:18
JOHN
676
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 at 8:18
JOHN
676
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
linear-algebra
asked Dec 1 at 8:18
JOHN
676
asked Dec 1 at 8:18
JOHN
676
asked Dec 1 at 8:18
JOHN
676
asked Dec 1 at 8:18
JOHN
676
asked Dec 1 at 8:18
JOHN
676
676
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 at 8:47
Quaternion
1966
add a comment |
up vote
5
down vote
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
add a comment |
up vote
2
down vote
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 at 9:37
Shubham Johri
2,502413
add a comment |
up vote
0
down vote
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 at 8:25
Theoop
1
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021116%2fneed-help-on-understanding-the-solution-linear-algebra-done-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 at 8:47
Quaternion
1966
add a comment |
up vote
3
down vote
accepted
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 at 8:47
Quaternion
1966
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 at 8:47
Quaternion
1966
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 at 8:47
Quaternion
1966
answered Dec 1 at 8:47
Quaternion
1966
answered Dec 1 at 8:47
Quaternion
1966
answered Dec 1 at 8:47
Quaternion
1966
1966
add a comment |
add a comment |
up vote
5
down vote
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
add a comment |
up vote
5
down vote
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
add a comment |
up vote
5
down vote
up vote
5
down vote
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
answered Dec 1 at 8:26
Siong Thye Goh
97.7k1463116
97.7k1463116
add a comment |
add a comment |
up vote
2
down vote
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 at 9:37
Shubham Johri
2,502413
add a comment |
up vote
2
down vote
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 at 9:37
Shubham Johri
2,502413
add a comment |
up vote
2
down vote
up vote
2
down vote
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 at 9:37
Shubham Johri
2,502413
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 at 9:37
Shubham Johri
2,502413
answered Dec 1 at 9:37
Shubham Johri
2,502413
answered Dec 1 at 9:37
Shubham Johri
2,502413
answered Dec 1 at 9:37
Shubham Johri
2,502413
2,502413
add a comment |
add a comment |
up vote
0
down vote
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 at 8:25
Theoop
1
add a comment |
up vote
0
down vote
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 at 8:25
Theoop
1
add a comment |
up vote
0
down vote
up vote
0
down vote
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 at 8:25
Theoop
1
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 at 8:25
Theoop
1
answered Dec 1 at 8:25
Theoop
1
answered Dec 1 at 8:25
Theoop
1
answered Dec 1 at 8:25
Theoop
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021116%2fneed-help-on-understanding-the-solution-linear-algebra-done-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
