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Linear Algebra With Applications
Found in: Page 1
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

In Exercises 57 through 61, consider a quadratic form q on R3 with symmetric matrix A, with the given properties. In each case, describe the level surface q(x)=1 geometrically.

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

Therefore, the solution is a pair of parallel planes.

See the step by step solution

Step by Step Solution

Step 1: Determine the diagonalized form

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.


Since A is a diagonal matrix, thus the eigenvalues of A are the diagonal entries of A viz.,A and λ3 .

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 λ1>0, λ2=0,λ3=0.

Hence, the quadratic form

qx=xTAx=x y zA=λ100000000xyz=λ1x2

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

x=1λ1and x=1λ1,

Step 2: Draw the graph

The quadratic form q(x,y,z)=x2 is positive semi-definite with rank (A) = 1 and the level curve q(x,y,z)=1 is sketched below:

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