1. Not finding help here? Sign up for a free 30min 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!

Invariant lines again

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Find all invariant lines, of the form y=mx for the matrix transformation.

    a) [itex] \left(
    \begin{array}{cc}
    5 & 15 \\
    -2 & 8
    \end{array}
    \right)[/itex]

    b) [itex]
    \left(
    \begin{array}{cc}
    3 & -5 \\
    -4 & 2
    \end{array}
    \right)[/itex]

    3. The attempt at a solution

    [itex] \left(
    \begin{array}{cc}
    5 & 15 \\
    -2 & 8
    \end{array}
    \right)\left(
    \begin{array}{c}
    x \\
    \text{mx}
    \end{array}
    \right)=\left(
    \begin{array}{c}
    5x+15\text{mx} \\
    -2x-8\text{mx}
    \end{array}
    \right)[/itex]

    And for it to map onto the line, this is probably the wrong bit.

    [itex]
    \left(
    \begin{array}{c}
    5x+15\text{mx} \\
    -2x-8\text{mx}
    \end{array}
    \right)=\left(
    \begin{array}{c}
    x \\
    \text{mx}
    \end{array}
    \right)[/itex]


    [itex]x=5x+15mx \Rightarrow 4=15m \Rightarrow m=\frac{4}{15}[/itex]

    I'll stop there because I know it's wrong. So what am I doing wrong, and what false assumptions have been made?

    I think if I put y=m(5x+15mx) i'll get the answer but I dont see the logic behind that at all even though it's been explained to be, in my head you could be going y=m(m(m(m(m(m(m(m... forever if you did that.

    Edit: Nope didn't work.
     
  2. jcsd
  3. Jun 16, 2009 #2

    Mark44

    Staff: Mentor

    I think that what you are asked to do here is to find all vectors (x, y) for which multiplying on the left by your matrix produces a new vector (kx, ky) that is a multiple of the input vector.

    I don't believe that there are any for your first matrix, but I believe there are two for your second matrix.

    If my understanding of this problem is correct, it is setting you up for a discussion of eigenvectors. An eigenvector is a nonzero vector x such that Ax = kx. IOW, multiplication of x by the matrix A produces a vector that is a scalar multiple of the input vector; i.e., a vector that has the same direction (possibly pointing the other direction) and therefore, has the same slope.
     
  4. Jun 16, 2009 #3
    error in the question, It supposed to be {{5,15},{-2,-8}} not +8!

    edit: yes you're right it's the preamble to the chaper on eigenvectors and eigenvalues, diagonalisation etc. It is still troubling me though, how I can find lines that map to themselves (invariant lines) such as y=mx or y=mx+c, but eigenvectors are fine. It's just this that's confusing me. An example would be good. Or the solution to these?
     
  5. Jun 16, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Saying a line is invariant does NOT mean that points are invariant. The line y= mx is invariant as long as a point, (x, mx) is mapped into something of the form (a, ma), not necessarily with a= x.
    You do not need
    [itex]\left(\begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx}\end{array}\right)=\left(\begin{array}{c} x \\ \text{mx}\end{array}\right)[/itex]
    rather you need
    [itex]\left(\begin{array}{c} 5x+15\text{mx} \\ -2x-8\text{mx}\end{array}\right)=\left(\begin{array}{c} a \\ \text{ma}\end{array}\right)[/itex]
    for some number a. More specifically, you need to look at
    [tex]\frac{-2x-8mx}{5x+ 15mx}= \frac{-2-8m}{5+ 15m}= \frac{ma}{a}= m[/tex]
    Solve that for m.
     
  6. Jun 16, 2009 #5

    Ok so,

    [itex]
    \left(
    \begin{array}{cc}
    5 & 15 \\
    -2 & -8
    \end{array}
    \right)\left(
    \begin{array}{c}
    x \\
    \text{mx}
    \end{array}
    \right)=\left(
    \begin{array}{c}
    5x+15\text{mx} \\
    -2x-8\text{mx}
    \end{array}
    \right)=\left(
    \begin{array}{c}
    x' \\
    \text{mx}'
    \end{array}
    \right)[/itex]

    [itex]mx'=m(5x+15mx)[/itex]

    [itex]-2x-8mx=m(5x+15mx)[/itex]

    [itex]\Rightarrow 15m^2+13m+2=0[/itex]

    [itex]m=-\frac{1}{5},-\frac{2}{3}[/itex]
     
  7. Jun 16, 2009 #6

    Mark44

    Staff: Mentor

    For your 2nd problem, calculate Ax, with x = (-5/4, 1)T. You should get a vector that is a constant multiple of x. In fact, when you multiply Ay, where y is any multiple of the vector I showed for x, you'll get another vector that is the same constant multiple of y. This shows that for this matrix A, the vector (-5/4, 1) is special in the sense that the product of A and this vector is another vector whose only difference is that it is longer than (-5/4, 1). For this matrix, there is another vector that behaves the same way.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Invariant lines again
  1. Matrix invariance (Replies: 1)

  2. Invariant lines? (Replies: 1)

  3. Trig again (Replies: 2)

  4. Probability again (Replies: 3)

Loading...