Answer to Question #108260 in Linear Algebra for Kentse

Question #108260
Find the entries xi;i D 1; 2; 3; 4 in the matrix P so that P is an orthogonal matrix where
P D
1
2
2
6
6
4
1 1 1 1
1 1
1
Expert's answer
2020-04-07T13:01:16-0400

Let matrix PP be


P=k(1abcdex1x211x3x41111)P=k\begin{pmatrix} 1 & a & b & c \\ d & e & x_1 & x_2 \\ 1 & 1 & x_3 & x_4 \\ 1 & 1 & 1 & 1 \end{pmatrix}

Let


r1=(1d11),r2=(ae11),r3=(bx1x31),r4=(cx2x41)r_1=\begin{pmatrix} 1 \\ d \\ 1 \\ 1 \end{pmatrix},r_2=\begin{pmatrix} a \\ e \\ 1 \\ 1 \end{pmatrix},r_3=\begin{pmatrix} b \\ x_1 \\ x_3 \\ 1 \end{pmatrix}, r_4=\begin{pmatrix} c \\ x_2 \\ x_4 \\ 1 \end{pmatrix}

 PP is an orthogonal matrix


PPT=IPP^T=I

Then


r1r2=r1r3=r1r4=r2r3=r2r4=r3r4=0r_1r_2=r_1r_3=r_1r_4=r_2r_3=r_2r_4=r_3r_4=0a+ed+1+1=0a+ed+1+1=0b+dx1+x3+1=0b+dx_1+x_3+1=0c+dx2+x4+1=0c+dx_2+x_4+1=0ab+ex1+x3+1=0ab+ex_1+x_3+1=0ac+ex2+x4+1=0ac+ex_2+x_4+1=0bc+x1x2+x3x4+1=0bc+x_1x_2+x_3x_4+1=0

Let d=1.d=1. Then

a+e+2=0a+e+2=0b+x1+x3+1=0b+x_1+x_3+1=0c+x2+x4+1=0c+x_2+x_4+1=0ab+ex1+x3+1=0ab+ex_1+x_3+1=0ac+ex2+x4+1=0ac+ex_2+x_4+1=0bc+x1x2+x3x4+1=0bc+x_1x_2+x_3x_4+1=0

Solve


a=1,b=1,c=1,d=1,e=1,a=-1, b=-1,c=1, d=1, e=-1,

x1=1,x2=1,x3=1,x4=1x_1=1, x_2=-1,x_3=-1, x_4=-1


k2r12=k2r22=k2r32=k2r42=1k^2r_1^2=k^2r_2^2=k^2r_3^2=k^2r_4^2=1

k=±0.5k=\pm0.5


P=0.5(1111111111111111),PT=0.5(1111111111111111)P=0.5\begin{pmatrix} 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix},P^T=0.5\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ 1 & 1 & -1 & 1 \end{pmatrix}

PPT=0.5(1111111111111111)(1111111111111111)=PP^T=0.5\begin{pmatrix} 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ 1 & -1 & -1 & 1 \end{pmatrix}=

=(1000010000100001)=I=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}=I

x1=1,x2=1,x3=1,x4=1x_1=1, x_2=-1,x_3=-1, x_4=-1



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