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Consider the 42 matrix

A=110[111111-1-11-11-11-1-11][20010000][3-443]

Use the result of Exercise 17 to find the least-squares solution of the linear system

Ax=bwhereb=[1234]

Work with paper and pencil.

Short Answer

Expert verified

x*=-110-165

Step by step solution

01

To find the least-squares solution of the linear system

Given that,

A=110111111-1-11-11-11-1-11200100003-443

Therefore, we have n=4,m=2,1=2,2=1. Also,

role="math" localid="1664872444698" u1=110110110110,u2=110110-110-110,v1=3-4,v2=43

Now the given linear system is:

Ax=1234

Thusb=1234

02

To find the least-squares solution of the linear system

Therefore, using the result of Exercise 17 we get that the least squares solution of the given linear system is given by:

x*=bu11v1+bu22v2=110+210+310+41023-4+110+210-310-410143=32-1610-2-1210=-110-165

03

Final proof

Thus, the required least squares solution of the given linear system is x*=-110-165.

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Most popular questions from this chapter

Let A be a22 matrix and ua unit vector in2. Show that2Au1,

where1,2 are the singular values of A. Illustrate this inequality with a sketch, and justify it algebraically.

Matrix [-2111-2111-2]is negative definite.

A Cholesky factorization of a symmetric matrix A is a factorization of the formA=LLTwhere L is lower triangular with positive diagonal entries.

Show that for a symmetricnnmatrix A, the following are equivalent:

(i) A is positive definite.

(ii) All principal submatricesrole="math" localid="1659673584599" A(m)of A are positive definite. See

Theorem 8.2.5.

(iii)det(Am)>0form=1,....,n

(iv) A has a Cholesky factorization A=LLT

Hint: Show that (i) implies (ii), (ii) implies (iii), (iii) implies (iv), and (iv) implies (i). The hardest step is the implication from (iii) to (iv): Arguing by induction on n, you may assume that A(n-1)) has a Cholesky factorization A(n-1)=BBT. Now show that there exist a vector xinRn-1and a scalar t such that

A=[An-1vvTk]=[B0xT1][BTx0t]

Explain why the scalar t is positive. Therefore, we have the Cholesky factorization

A=[B0xTt][BTx0t]

This reasoning also shows that the Cholesky factorization of A is unique. Alternatively, you can use the LDLT factorization of A to show that (iii) implies (iv).See Exercise 5.3.63.

To show that (i) implies (ii), consider a nonzero vector, and define

role="math" localid="1659674275565" y=[x0M0]

In Rn(fill in n 鈭 m zeros). Then

role="math" localid="1659674437541" xA(m)x=yTAy>0

For the quadratic form q(x1x2)=8x12-4x1x2+5x22, find an orthogonal basis w1,w2of R2such that q(c1w1+c2w2)=c12+c22. Use your answer to sketch the level curve q(x~)=1. Compare with Example 4 and Figure 4 in this section. Exercise 63 is helpful.

Sketch the curves defined in Exercises 15 through 20. In each case, draw and label the principal axes, label the intercepts of the curve with the principal axes, and give the formula of the curve in the coordinate system defined by the principal axes.

18.9x124x1x2+6x22=1

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