91影视

Consider the positive definite quadratic form q(x, y, z) , which is determined by a symmetric matrix B. We know that the quadratic form q is diagonalizable with regard to the orthonormal eigenbasis of B because of Theorem 8.2.2.

With the diagonal matrix, consider the diagonalized form of q.

$A=\left[\begin{array}{ccc}{位}_1& 0& 0\\ 0& {位}_2& 0\\ 0& 0& {位}_3\end{array}\right]$

Because A is a diagonal matrix, the eigenvalues of A are the diagonal entries of A, which are ${位}_1,{位}_2\mathrm{and}{位}_3$.

We know that a matrix is positive definite (positive semi-definite) if and only if all of its eigenvalues are positive because of Theorem 8.2.4. (non-negative). A symmetric matrix X is also called negative definite (semi-definite) if and only if X is called positive definite (positive semi-definite). As a result, we have negative eigenvalues for all eigenvalues of a negative definite (semi-definite) matrix (non-positive).

Because A is positive definite, all of the eigenvalues must be positive, and so

${位}_1>0,{位}_2>0,{位}_3>0$

As a result, the quadratic form is used.

$$\begin{gathered} \mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}=\left[\mathrm{x}\mathrm{y}\mathrm{z}\right]\left[\begin{array}{ccc}{\mathrm{位}}_1& 0& 0\\ 0& {\mathrm{位}}_2& 0\\ 0& 0& {\mathrm{位}}_3\end{array}\right]\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right] \\ ={\mathrm{位}}_1{\mathrm{x}}^2+{\mathrm{位}}_2{\mathrm{y}}^2+{\mathrm{位}}_3{\mathrm{z}}^2 \end{gathered}$$

Is positive definite and the level curve $\mathrm{q}\left(\mathrm{x}\right)=1$represents an ellipsoid.

The quadratic form $\mathrm{q}(\mathrm{x},\mathrm{y},\mathrm{z})==0.1{\mathrm{x}}^2+0.2{\mathrm{y}}^2+0.3{\mathrm{z}}^2$is positive definite and the level curve $\mathrm{q}\left(\mathrm{x}\right)=1$, that is $0.1{\mathrm{x}}^2+0.2{\mathrm{y}}^2+0.3{\mathrm{z}}^2=1$is an ellipsoid, sketched below:

The final solution is ellipsoid