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!

Linear Algebra - Gram-Schmidt Process

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Let R2 have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.

    u1 = (1,-3)
    u2 = (2,2)

    2. Relevant equations

    Gram-Schmidt process:

    [tex]
    \\v_1 = u_1
    \\v_2= u_2 -

    \frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \|
    v_1

    \right \| ^ 2}


    v_1
    [/tex]

    3. The attempt at a solution

    We'll go ahead and find the norm of v first. Then we'll evaluate the second Gram-Schmidt equation.

    [tex]
    \\\left \| v_1 \right \| = \sqrt{10} \\\\
    \left \| v_1 \right \|v_1 = u_1 = (1,-3)
    \\\\\left \| v_2 \right \|v_2= \begin{bmatrix}
    2\\2

    \end{bmatrix} -

    \frac{\left ( \left \langle \begin{bmatrix}
    2\\2

    \end{bmatrix} , \begin{bmatrix}
    \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

    \end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix}
    \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

    \end{bmatrix} \right \| ^ 2}

    \begin{bmatrix}
    \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

    \end{bmatrix}


    \\\\\left \| v_2 \right \|v_2= \begin{bmatrix}
    2\\2

    \end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix}

    \\\\

    \begin{bmatrix}
    2\\2

    \end{bmatrix}

    - \begin{bmatrix}
    \frac{-4}{10}\\\\\frac{12}{10}

    \end{bmatrix}

    =

    \begin{bmatrix}
    \frac{24}{10}\\\\\frac{8}{10}

    \end{bmatrix}

    =

    \begin{bmatrix}
    \frac{12}{5}\\\\\frac{4}{5}

    \end{bmatrix}
    [/tex]

    I've not been able to get another answer for this step. Even checked in Matlab.

    Now, we'll normalize the vector like the problem wants us to:
    [tex]
    \\\left \| v_2 \right \| = \sqrt{8}
    \\
    v_2 = \begin{bmatrix}
    \frac{\frac{12}{5}}{\sqrt{8}}\\
    \frac{\frac{4}{5}}{\sqrt{8}}

    \end{bmatrix}

    =

    \begin{bmatrix}

    \frac{\frac{12}{5}}{\sqrt{8}}\\
    \frac{\frac{4}{5}}{\sqrt{8}}

    \end{bmatrix}

    =

    \begin{bmatrix}

    \frac{12}{10\sqrt{2}}\\\\
    \frac{4}{10\sqrt{2}}

    \end{bmatrix}

    =

    \begin{bmatrix}

    \frac{6}{5\sqrt{2}}\\\\
    \frac{2}{5\sqrt{2}}

    \end{bmatrix}

    =

    \begin{bmatrix}

    \frac{6\sqrt{2}}{10}\\\\
    \frac{2\sqrt{2}}{10}

    \end{bmatrix}

    =

    \begin{bmatrix}

    \frac{3\sqrt{2}}{5}\\\\
    \frac{\sqrt{2}}{5}

    \end{bmatrix}
    [/tex]

    This is an odd problem, so I know that the solution should be [itex] v_2 = (\frac{3}{\sqrt{10}} , \frac{1}{\sqrt{10}}) [/itex]. There must be something wrong when I'm finding [itex]||v_2||v_2[/itex] . Is there anything fundamentally wrong with my process?
     
    Last edited: Apr 15, 2016
  2. jcsd
  3. Apr 15, 2016 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    If you get rid of a common factor of 4/5, you have ##\vec{v}_2 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}##.

     
  4. Apr 17, 2016 #3
    Ok. So I gather that my answer is not technically wrong? How can I verify that my original answer is in fact an orthonormal basis spanning the same subspace?

    I guess if I make computational errors, it will be unlikely that all the dot products in some generated set of vectors is zero. Right?
     
    Last edited: Apr 17, 2016
  5. Apr 18, 2016 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your answer is wrong. You didn't normalize the vector correctly.
     
  6. Apr 18, 2016 #5
    Ah, I see that I accidentally used the norm of u2 instead of v2 to normalize v2. The book confused me by switching up u's and v's.

    Checking to see if the norm of the final vectors is one would also be a useful trick.
     
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: Linear Algebra - Gram-Schmidt Process
  1. Linear Algebra (Replies: 5)

  2. Linear Algebra (Replies: 1)

Loading...