Answer to Question #157252 in Linear Algebra for Emerald

Question #157252

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. 


1
Expert's answer
2021-01-22T11:50:55-0500

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


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


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

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


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


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


Swap second and third rows:


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


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

"s_1-2s_2+s_3+s_4=-1 \\\\\n7s_2-2s_3-2s_4=4 \\\\\n3s_3+s_4=0 \\\\\n2s_4=0"


"s_1-2s_2+s_3=-1 \\\\\n7s_2-2s_3=4 \\\\\n3s_3=0 \\\\\ns_4=0"


"s_1-2s_2=-1 \\\\\n7s_2=4 \\\\\ns_3=0 \\\\\ns_4=0"


"s_1-8\/7=-1 \\\\\ns_2=4\/7 \\\\\ns_3=0 \\\\\ns_4=0"


"s_1=1\/7 \\\\\ns_2=4\/7 \\\\\ns_3=0 \\\\\ns_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)=\\\\\n=1\/7(1-8, -1+8, 2+12, -1+8)=\\\\\n=1\/7(-7, 7, 14, 7)=(-1,1,2,1)"


Solution is correct.


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