Extreme Value Problem involving an Ellipse and Lagrange Multipliers

Question

I`m trying to solve a given question, "Let $2L$ denote the length of the longest axis of the ellipsoid $$ax^2+by^2+cz^2+2dxy+2exz+2fyz=1.$$ Prove that $L$ is the greatest real root of the following equation in $\lambda$: $$\text{det}=\begin{bmatrix} a-\frac{1}{\lambda^2} & d & e \\ d & b-\frac{1}{\lambda^2} & f \\ e & f & c-\frac{1}{\lambda^2} \end{bmatrix}=0."$$ I first began by letting $A\in M_{3\times3}(\mathbb{R})$ be symmetric and $f(v)=(Av)\cdot v$, where $v\in\mathbb{R}^3$. I also know that Lagrange multipliers are to be used somewhere along the proof, but I'm unaware of where to go from here. However, I would think that if I were to do this using lagrange multipliers, my constraint would have to be $$g(x,y,z)=ax^2+by^2+cz^2+2dxy+2exz+2fyz-1.$$ Help is appreciated.

Answer 1

We can start by observing that $$ 1=ax^2+by^2+cz^2+2dxy+2exz+2fyz=\\ [x\ y\ z]\left[\array{a&d&e\\d&b&f\\e&f&c}\right]\left[\array{x\\y\\z}\right].\quad\quad (1)$$

Let the matrix above be denoted by $A$. Note that the matrix A is symmetric. Hence it has a decomposition of$$ A=U^T\Lambda U,$$ where $U$ is unitary matrix, and $\Lambda=\left[\array{\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3}\right]$ is the diagonal matrix of eigenvalues. By the definition of eigenvalues. $\lambda_1, \lambda_2, \lambda_3$ are roots of equation $$\text{det}\begin{bmatrix} a-\lambda & d & e \\ d & b-\lambda & f \\ e & f & c-\lambda \end{bmatrix}=0. \quad\quad(2) $$

Now (1) can be rewritten as $$ \\ [x\ y\ z]U^T\Lambda U\left[\array{x\\y\\z}\right] = [x^\prime\ y^\prime\ z^\prime]\Lambda \left[\array{x^\prime\\y^\prime\\z^\prime}\right],\quad\quad (3)$$ where $\left[\array{x^\prime\\y^\prime\\z^\prime}\right]=U\left[\array{x\\y\\z}\right].$ Recall that $U$ is a unitary matrix, hence the new coordinate system is a rotation of the old coordinate system. And the ellipsoid described by $x^\prime,y^\prime,z^\prime$ has the same shape (and axis sizes) as the original ellipsoid. We can further rewrite (3) as $$ \lambda_1{x^\prime}^2 + \lambda_2{y^\prime}^2 + \lambda_3{z^\prime}^2 = 1.\quad\quad (4) $$ From (4), we can tell, the $L=\max\left\{\sqrt{1/\lambda_1}, \sqrt{1/\lambda_2}, \sqrt{1/\lambda_3}\right\} $ (half of the longest axis length). Without loss of generality, let's assume $\lambda_1 \le \lambda_2, \lambda_1\le\lambda_3$. Then $L=\sqrt{1/\lambda_1}$, and $\lambda_1=\frac{1}{\lambda^2}$. Recall that $\lambda_1$ is a root (the smallest) of equation (2). Hence $\lambda$ is the largest root of $$ \text{det}\begin{bmatrix} a-\frac{1}{\lambda^2} & d & e \\ d & b-\frac{1}{\lambda^2} & f \\ e & f & c-\frac{1}{\lambda^2} \end{bmatrix}=0.$$