Answer to Question #108260 in Linear Algebra for Kentse

Question

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

Answer 1

Let matrix $P$ be

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

Let

$$r_1=\begin{pmatrix} 1 \\ d \\ 1 \\ 1 \end{pmatrix},r_2=\begin{pmatrix} a \\ e \\ 1 \\ 1 \end{pmatrix},r_3=\begin{pmatrix} b \\ x_1 \\ x_3 \\ 1 \end{pmatrix}, r_4=\begin{pmatrix} c \\ x_2 \\ x_4 \\ 1 \end{pmatrix}$$

 $P$ is an orthogonal matrix

$$PP^T=I$$

Then

$$r_1r_2=r_1r_3=r_1r_4=r_2r_3=r_2r_4=r_3r_4=0$$ $$a+ed+1+1=0$$ $$b+dx_1+x_3+1=0$$ $$c+dx_2+x_4+1=0$$ $$ab+ex_1+x_3+1=0$$ $$ac+ex_2+x_4+1=0$$ $$bc+x_1x_2+x_3x_4+1=0$$

Let $d=1.$ Then

$$a+e+2=0$$ $$b+x_1+x_3+1=0$$ $$c+x_2+x_4+1=0$$ $$ab+ex_1+x_3+1=0$$ $$ac+ex_2+x_4+1=0$$ $$bc+x_1x_2+x_3x_4+1=0$$

Solve

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

$$x_1=1, x_2=-1,x_3=-1, x_4=-1$$

$$k^2r_1^2=k^2r_2^2=k^2r_3^2=k^2r_4^2=1$$

$$k=\pm0.5$$

$$P=0.5\begin{pmatrix} 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix},P^T=0.5\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ 1 & 1 & -1 & 1 \end{pmatrix}$$

$$PP^T=0.5\begin{pmatrix} 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ 1 & -1 & -1 & 1 \end{pmatrix}=$$

$$=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}=I$$

$x_1=1, x_2=-1,x_3=-1, x_4=-1$