Question

Given the vector u = (−1, 1, 2, 1) and the vectors v1 = (1, −1, 2, −1), v2 =

(−2, 2, 3, 2), v3 = (1, 2, 0, −1), v4 = (1, 0, 0, 1) which form an orthogonal ba-

sis for IR 4

, find the coordinate vector (u)S for the vector u. 

Answer 1

Coordinates of u in basis v1, v2, v3, v4 are numbers s1, s2, s3, s4, such as:

$u=s_1v_1+s_2v_2+s_3v_3+s_4v_4$

We can write this using matrix:

$\begin{pmatrix} v_{11} & v_{21} & v_{31} & v_{41} \\ v_{12} & v_{22} & v_{32} & v_{42} \\ v_{13} & v_{23} & v_{33} & v_{43} \\ v_{14} & v_{24} & v_{34} & v_{44} \\ \end{pmatrix} \begin{pmatrix} s_1 \\ s_2 \\ s_3 \\ s_4 \end{pmatrix} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{pmatrix}$

We've got the linear equation. We can solve it using matrix transformations (Gaussian method):

$\begin{pmatrix} v_{11} & v_{21} & v_{31} & v_{41} &|& u_1\\ v_{12} & v_{22} & v_{32} & v_{42} &|& u_2 \\ v_{13} & v_{23} & v_{33} & v_{43} &|& u_3 \\ v_{14} & v_{24} & v_{34} & v_{44} &|& u_4 \\ \end{pmatrix} =$

$= \begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ -1 & 2 & 2 & 0 &|& 1 \\ 2 & 3 & 0 & 0 &|& 2 \\ -1 & 2 & -1 & 1 &|& 1 \\ \end{pmatrix}$

Add first row to the second and fourth, and subtract 2 times first row from third row:

$\begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ 0 & 0 & 3 & 1 &|& 0 \\ 0 & 7 & -2 & -2 &|& 4 \\ 0 & 0 & 0 & 2 &|& 0 \\ \end{pmatrix}$

Swap second and third rows:

$\begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ 0 & 7 & -2 & -2 &|& 4 \\ 0 & 0 & 3 & 1 &|& 0 \\ 0 & 0 & 0 & 2 &|& 0 \\ \end{pmatrix}$

After transformations, we've got this system of equations:

$s_1-2s_2+s_3+s_4=-1 \\ 7s_2-2s_3-2s_4=4 \\ 3s_3+s_4=0 \\ 2s_4=0$

$s_1-2s_2+s_3=-1 \\ 7s_2-2s_3=4 \\ 3s_3=0 \\ s_4=0$

$s_1-2s_2=-1 \\ 7s_2=4 \\ s_3=0 \\ s_4=0$

$s_1-8/7=-1 \\ s_2=4/7 \\ s_3=0 \\ s_4=0$

$s_1=1/7 \\ s_2=4/7 \\ s_3=0 \\ s_4=0$

Check the solution. If solution is correct, we'll get (-1, 1, 2, 1):

$u=1/7v_1+4/7v_2=1/7(v_1+4v_2)=\\ =1/7(1-8, -1+8, 2+12, -1+8)=\\ =1/7(-7, 7, 14, 7)=(-1,1,2,1)$

Solution is correct.