1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basis and Components (Vectors)

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ##u_1,u_2,u_3## be a basis and let ##v_1=-u_1+u_2-u_3## , ## v_2=u_1+2u_2-u_3## , ##v_3=2u_1+u_3## show that ##v_1,v_2,v_3## is a basis and find the components of ##a=2u_1-u_3## in terms of ##v_1,v_2,v_3##

    2. Relevant equations
    For basis vecor we need to clarify these vector are lineerly independet.We can understand linearly independent as make a matrix 3x3 and see the take the det of matrx. If det of matrix is not zero means linearly independent.
    3. The attempt at a solution
    I take the determinant of ##v_1,v_2,v_3## and I found -1.It means linearly independent.It means they are basis vectors.Then I tried to show ##u_1=(1,0,0)## then I tried write it in terms of ##v## but I thought that there a lot way to do that or maybe wrong.

    The answer is ##u_1=-2v_1+v_2-v_3## and ##u_2=3v_1-v_2+2v_3## and ##u_3=4v_1-2v_2+3v_3## and ##a=-8v_!+4v_2-5v_3##
    Thanks
     
  2. jcsd
  3. Jun 11, 2015 #2

    ChrisVer

    User Avatar
    Gold Member

    The matrix you are looking the determinant of is:
    [itex] \begin{bmatrix} -1 & 1 & -1 \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{bmatrix}[/itex]
    Which is -1.

    Now you have to go the other way around. I mean you know how [itex]v[/itex]'s are given in terms of [itex]u[/itex]'s... you should actually go to the inverse relation and see how [itex]u[/itex]'s are written in terms of [itex]v[/itex]'s (so you can make the substitution in alpha)...

    If you are lucky you can see what you have to do right away by a few tries of adding/substracting and multiplying with factors the [itex]v_i[/itex].

    Try to find [itex]u_1[/itex] in terms of [itex]v_{1,2,3}[/itex].

    (do you know about inverse matrices?)
     
  4. Jun 11, 2015 #3
    Yeah,I know.
     
  5. Jun 11, 2015 #4

    ChrisVer

    User Avatar
    Gold Member

    you can use the inverse matrix to get the [itex]u_i[/itex] in terms of [itex]v_j[/itex]:

    [itex] \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 2 & -1 \\ 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} [/itex]

    To find ##M## here:

    [itex] \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}= M \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} [/itex]
     
  6. Jun 11, 2015 #5

    ChrisVer

    User Avatar
    Gold Member

    You can try also your way [itex] u_1 = \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}[/itex], [itex] u_2 = \begin{pmatrix} 0 \\ 1 \\0 \end{pmatrix}[/itex], [itex] u_3 = \begin{pmatrix} 0\\ 0 \\1 \end{pmatrix}[/itex].

    That means you are taking a particular basis for [itex]u_i[/itex]. Then the [itex]v[/itex]'s are:

    [itex] v_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}[/itex], [itex] v_2 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}[/itex], [itex] v_3 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}[/itex].

    Now you would like to see how you can add [itex]v_1,v_2,v_3[/itex] in order to get [itex]u_1[/itex]. You can try that by solving the equation for [itex]a,b,c[/itex]:

    [itex]u_1 = av_1+ bv_2 +cv_3 \Rightarrow \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}= a \begin{pmatrix} -1 \\ 1 \\-1 \end{pmatrix}+ b\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} +c \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}[/itex].
    That's 3 equations with 3 unknown variables (a,b,c) so it's solvable:
    [itex] \begin{align*} -a +b& +2c = 1 \\ a+2b&=0 \\ -a-b&+c =0 \end{align*} [/itex]

    And similarily work for the rest [itex] u_2 \& u_3[/itex]. That will be in total 9 equations with 9 unknown parameters.

    That's an alternative way of saying that you are taking the inverse matrix.. you just wrote for [itex]M= \begin{bmatrix} a & b & c \\ d & f & e \\ h & w & r \end{bmatrix}[/itex] and you go to determine its elements one by one.
     
    Last edited: Jun 11, 2015
  7. Jun 11, 2015 #6
    ##1=-a+b+2c##
    ##0=a+2b##
    ##0=-a-b+c##
    then #### ##a=-2b## then If I put this in third equation equation I found ##b=-c## .If I put these info on first equation I get ##1=2b+b-2b##, ##b=1## then ##a=-2## and ##c=-1## Which İt fits the answer.
    I understand the matrix way and this way.Matrix way is look like simpler but Its hard to find inverse matrix of 3x3.I used online calculator and I found exact solution.
    Thanks my friend.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Basis and Components (Vectors)
  1. Basis of vector (Replies: 10)

  2. Basis vector (Replies: 8)

  3. Component of a vector (Replies: 1)

  4. Basis of a Vector Space (Replies: 15)

Loading...