Results 1 to 5 of 5

Math Help - Linearly Independent

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    71

    Linearly Independent

    I am having difficulties with the following problem. I need to find values of t that make the set linearly independent.

    Problem:
     S = {(t,1,1), (1,t,1), (1,1,t)}

    The answer is all t not equal to 1 and -2. I am confused on how to approach this problem.

    If I set up a matrix I believe I see where -2 comes from. And we need t to not equal that otherwise that's another solution to the homogeneous equation. Matrix:


    \left(\begin{array}{cccc}t&1&1&0\\1&t&1&0\\1&1&t&0  \end{array}\right)

    Beyond seeing where those two numbers come from, I am not sure of the procedure behind getting them. Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Alterah View Post
    I am having difficulties with the following problem. I need to find values of t that make the set linearly independent.

    Problem:
     S = {(t,1,1), (1,t,1), (1,1,t)}

    The answer is all t not equal to 1 and -2. I am confused on how to approach this problem.

    If I set up a matrix I believe I see where -2 comes from. And we need t to not equal that otherwise that's another solution to the homogeneous equation. Matrix:


    \left(\begin{array}{cccc}t&1&1&0\\1&t&1&0\\1&1&t&0  \end{array}\right)

    Beyond seeing where those two numbers come from, I am not sure of the procedure behind getting them. Thanks for any help.

    Erase that last column of zeroes (it serves no purpose in this case), and bring your matrix to echelon form, perhaps first interchanging the first and second rows, so that you'll have 1 in the first entry...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    71
    Quote Originally Posted by tonio View Post
    Erase that last column of zeroes (it serves no purpose in this case), and bring your matrix to echelon form, perhaps first interchanging the first and second rows, so that you'll have 1 in the first entry...

    Tonio
    If I try bringing it to RREF, I get a jumble for a matrix. Taking your advice:

    \left(\begin{array}{ccc}1&t&1\\t&1&1\\1&1&t\end{ar  ray}\right)<br />
    \left(\begin{array}{ccc}1-t&t-1&0\\t-1&0&1-t\\0&1-t&t-1\end{array}\right)<br />

    If t = 1 here I get the trivial solution as everything becomes zero. I still think I am missing something. Not seeing where the -2 is coming from regarding the above matrix. What I have done seems to imply t = 2. Thanks for your patience.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2009
    Posts
    153
    if \det(A)\neq 0 matrix A is invertible,then row and column vectors are linear independence
    So find f(t)=\det\begin{bmatrix}t&1&1\\1&t&1\\1&1&t\end{bm  atrix}=\det\begin{bmatrix}0&1-t^2&1-t\\0&t-1&1-t\\1&1&t\end{bmatrix}=\det\begin{bmatrix}1-t^2&1-t\\t-1&1-t\end{bmatrix}=t^3-3t+2

    if f(t)=0 \rightarrow \{t=1,t=-2\}, then matrix A is noninvertible, therefore S linearly dependent

    So When t\neq 1\ or\ -2\ , \ S is linearly independent
    Last edited by math2009; November 2nd 2009 at 03:56 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,567
    Thanks
    1409
    Quote Originally Posted by Alterah View Post
    I am having difficulties with the following problem. I need to find values of t that make the set linearly independent.

    Problem:
     S = {(t,1,1), (1,t,1), (1,1,t)}

    The answer is all t not equal to 1 and -2. I am confused on how to approach this problem.

    If I set up a matrix I believe I see where -2 comes from. And we need t to not equal that otherwise that's another solution to the homogeneous equation. Matrix:


    \left(\begin{array}{cccc}t&1&1&0\\1&t&1&0\\1&1&t&0  \end{array}\right)

    Beyond seeing where those two numbers come from, I am not sure of the procedure behind getting them. Thanks for any help.
    I would be inclined to go back to the basic definition of "linearly independent": for what t does a(t,1,1)+ b(1,t,1)+ c(1,1,t)= (0,0,0) have only the solution a= b= c= 0? That gives the three equations at+b+c= 0, a+bt+ c= 0, and a+ b+ ct= 0 (Which, of course have exactly the coefficient matrix you use). If we subtract the second equation from the first, we get (t-1)a+ (1-t)b= 0 or (t-1)a= (t-1)b so a= b for all t except t= 1. Similarly, subtracting the third matrix from the first gives (t- 1)a+ (1- t)c= 0 or (t-1)a= (t-1)c so a= c for all t except 1. Putting b=a and c= a into the first equation, ta+ a+ a= (t+2)a= 0 and we must have a= b= 0 unless t= -2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linearly dependent in Q, linearly independent in R
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: April 12th 2011, 01:44 PM
  2. linearly independent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 17th 2010, 11:55 AM
  3. linearly independent
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 14th 2010, 08:09 AM
  4. Linearly Independent?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 6th 2008, 02:40 PM
  5. linearly independent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 30th 2008, 05:02 AM

Search Tags


/mathhelpforum @mathhelpforum