Quadratic forms under constraints

[mr.tea]

Homework Statement

Find the minimum value of $x_1^2+x_2^2+x_3^2$ subject to the constraint:
$q(x_1,x_2,x_3)=7x_1^2+3x_2^2+7x_3^2+2x_1x_2+4x_2x_3=1$

Homework Equations

The Attempt at a Solution

I am not really sure how to think about it. I have seen the opposite way but have not seen this type of question yet. Any guidance will be very helpful.

Thank you.

 

[Ray Vickson]

Homework Statement

Find the minimum value of $x_1^2+x_2^2+x_3^2$ subject to the constraint:
$q(x_1,x_2,x_3)=7x_1^2+3x_2^2+7x_3^2+2x_1x_2+4x_2x_3=1$

Homework Equations

The Attempt at a Solution

I am not really sure how to think about it. I have seen the opposite way but have not seen this type of question yet. Any guidance will be very helpful.

Thank you.

What do you mean by "the opposite way"?

 

[mr.tea]

What do you mean by "the opposite way"?

I mean something like "find the max/min of $q(x_1,x_2,x_3)=7x_1^2+3x_2^2+7x_3^2+2x_1x_2+4x_2x_3$ under the constraint $x_1^2+x_2^2+x_3^2=1$"
I know how to solve this type of questions. This is the opposite way(maybe "way" is not a good word here)

 

[Samy_A]

I mean something like "find the max/min of $q(x_1,x_2,x_3)=7x_1^2+3x_2^2+7x_3^2+2x_1x_2+4x_2x_3$ under the constraint $x_1^2+x_2^2+x_3^2=1$"
I know how to solve this type of questions. This is the opposite way(maybe "way" is not a good word here)

How would you solve the "opposite way" problem? And why does that method not work in the present case?

 

[mr.tea]

How would you solve the "opposite way" problem? And why does that method not work in the present case?

Because there is a theorem that says that at the unit sphere($x_1^2+x_2^2+x_3^2=1$) the max/min of the equation is at the max/min of the eigenvalues of the form. I cannot use it here.

 

[Ray Vickson]

Because there is a theorem that says that at the unit sphere($x_1^2+x_2^2+x_3^2=1$) the max/min of the equation is at the max/min of the eigenvalues of the form. I cannot use it here.

Yes, you can.

You can change variables to $u_i = U_i(x_1,x_2,x_3)$ for $i = 1,2,3$, where the $U_i$ are linear functions of the $x_j$, devised so that your $q(x_1,x_2,x_3)$ has the form $u_1^2 + u_2^2 + u_3^2$. Then, if you reverse the transformations to get $x_i = X_i(u_1,u_2,u_3)$, the functions $X_i$ will be linear in the $u_j$, and so $f(x_1,x_2,x_3) = x_1^2 + x_2^2 + x_3^2$ will be quadratic in the $u_j$. That problem will be exactly of the type you can solve already.

However, I don't know why you would ever want to do this; it is much easier to just solve the problem directly using the Lagrange multiplier method.

 

[mr.tea]

Yes, you can.

You can change variables to $u_i = U_i(x_1,x_2,x_3)$ for $i = 1,2,3$, where the $U_i$ are linear functions of the $x_j$, devised so that your $q(x_1,x_2,x_3)$ has the form $u_1^2 + u_2^2 + u_3^2$. Then, if you reverse the transformations to get $x_i = X_i(u_1,u_2,u_3)$, the functions $X_i$ will be linear in the $u_j$, and so $f(x_1,x_2,x_3) = x_1^2 + x_2^2 + x_3^2$ will be quadratic in the $u_j$. That problem will be exactly of the type you can solve already.

However, I don't know why you would ever want to do this; it is much easier to just solve the problem directly using the Lagrange multiplier method.

Thank you for the answer. I am not sure that I got it 100% but I will work on it.
This is a question from a past exam in linear algebra. The subject of Lagrange multiplier is not covered in this course. I am not sure if I want to start talking about the course. I have a lot of bad things to say...

By the way, now I understand the reason why I couldn't find any explanation or notes about something similar to that...

Thank you for the help, I appreciate it!

Thomas