91影视

Consider the positive semi-definite quadratic form q ( x, y , z ) , which is described by a symmetric matrix B of rank (B) = 1 . We know that the quadratic form q is diagonalizable with regard to the orthonormal eigen basis of B because of Theorem 8.2.2.

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

Since A is a diagonal matrix, thus the eigenvalues of A are the diagonal entries of A viz.,A 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).

Recall that A is positive semi-definite with rank (A) =1 so, by the previous discussion, one eigenvalues must be positive and other two eigenvalues must be zero. Without loss of generality let${\mathrm{位}}_1>0,{\mathrm{位}}_2=0,{\mathrm{位}}_3=0.$

Hence, the quadratic form

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

is positive semi-definite and the level curve q (x) = 1 represents two parallel planes, viz.

$x=\frac{1}{\sqrt{{位}_1}}\mathrm{and}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{位}}_1}}$,

The quadratic form q(x,y,z)=$x^2$ is positive semi-definite with rank (A) = 1 and the level curve q(x,y,z)=1 is sketched below: