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Q59E

Expert-verifiedFound in: Page 1

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**In Exercises 57 through 61, consider a quadratic form q on ${{\mathbf{R}}}^{{\mathbf{3}}}$ with symmetric matrix A, with the given properties. In each case, describe the level surface ${\mathbf{q}}{\left(\stackrel{\rightharpoonup}{x}\right)}{\mathbf{=}}{\mathbf{1}}$ geometrically.**

**59.** q** is positive semidefinite and rank A = 1.**

Therefore, the solution is a pair of parallel planes.

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.

$\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{\lambda}}_{1}& 0& 0\\ 0& {\mathrm{\lambda}}_{2}& 0\\ 0& 0& {\mathrm{\lambda}}_{3}\end{array}\right]$

Since A is a diagonal matrix, thus the eigenvalues of A are the diagonal entries of A viz.,A and ${\lambda}_{3.}$

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 ${\mathrm{\lambda}}_{1}>0,{\mathrm{\lambda}}_{2}=0,{\mathrm{\lambda}}_{3}=0.$

Hence, the quadratic form

$\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}=\left[\mathrm{x}\mathrm{y}\mathrm{z}\right]\mathrm{A}=\left[\begin{array}{ccc}{\mathrm{\lambda}}_{1}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]={\mathrm{\lambda}}_{1}{\mathrm{x}}^{2}$

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

$x=\frac{1}{\sqrt{{\lambda}_{1}}}\mathrm{and}\mathrm{x}=\frac{1}{\sqrt{{\mathrm{\lambda}}_{1}}}$,

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:

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