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Proving determinant is null

  1. Jan 20, 2013 #1
    Hi, I'm tutoring someone on lineal algebra (upper high-school level).
    We're going through some exercises, but we're stuck with a trivial
    looking one.

    1. The problem statement, all variables and given/known data
    Show that the following determinant is null:

    \left| \begin{array}{ccc}
    1 & \cos x & \cos 2x \\
    \cos x & \cos 2x & \cos 3x \\
    \cos 2x & \cos 3x & \cos 4x \end{array} \right|

    2. Relevant equations

    $$\cos 2x = \cos^2 x - \sin^2 x$$
    $$\cos 3x = \cos^3 x - 3 \cos x \sin^2 x$$
    $$\cos 4x = \cos^4 x - 6 \cos^2 x \sin^2 x + \sin^4 x$$

    Expansion of a determinant by the elements of a row or column.

    3. The attempt at a solution

    I tried a brute force approach by expanding the cosines and computing the determinant directly in terms of $$\sin x$$ and $$\cos x$$, but that's an inelegant approach. There has to be a trivial, neater way to show this, but I seem to not be in my brightest lately, as I can't find it. I suspect I should be using the fact that the matrix is symmetric. Any ideas?
     
  2. jcsd
  3. Jan 20, 2013 #2

    Ray Vickson

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    There is an easier way, but maybe not completely suitable for an upper high-school student.

    First note that ##\cos(kx) = \text{Re}(e^{ikx})\:\: (i = \sqrt{-1})##, and consider the linear system of equations in real ##y_1,y_2, y_3:##
    [tex] y_1 + e^{ix} y_2 + e^{2ix} y_3 = 0\\
    e^{ix} y_1 + e^{2ix} y_2 + e^{3ix} y_3 = 0\\
    e^{2ix} y_1 + e^{3ix} y_2 + e^{4ix} y_3 = 0,
    [/tex]
    Taking the real parts gives the equations involving cos(kx).

    We see that the second equation is obtained from the first by multiplying both sides by ##e^{ix}##, and the third is obtained from the first by multiplying both sides by ##e^{2ix}##. Therefore, the second and third equations are just scalar multliples of the first one, so the system has coefficient matrix of rank 1, hence of zero determinant.
     
  4. Jan 20, 2013 #3
    Oh, of course, it was so obvious. Now I feel silly ;-)
    Thanks a lot for this, this is the sort of solution I was looking for (I agree, maybe better suitable for a college freshman than a high-schooler, but I think I can make it work for him).
    Again, thanks.
     
  5. Jan 20, 2013 #4

    Hurkyl

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    There's no reason to think Det(A + Bi) = 0 should imply Det(A) = 0.

    As a concrete example, consider the matrix

    [tex]\left( \begin{matrix} 1 & 0 \\ 1 & -1 \end{matrix} \right)[/tex]

    which is the real part of the matrix

    [tex]\left( \begin{matrix} 1 & i \\ i+1 & i-1 \end{matrix} \right)[/tex]
     
  6. Jan 20, 2013 #5

    Dick

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    I don't think that's an entirely legitimate argument. It's pretty easy to write down a complex matrix with zero determinant with a real part that has nonzero determinant. And looking at the case x=1 numerically I get that it's rank 2, not rank 1.
     
  7. Jan 20, 2013 #6

    Ray Vickson

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    Right. After posting the argument I did worry about that, and became much less confident in the original argument.
     
  8. Jan 20, 2013 #7

    LCKurtz

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    Heh. Welcome to the club Ray. I've done that more than once on these forums, sometimes in a very embarrassing fashion.
     
  9. Jan 20, 2013 #8

    I like Serena

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    Let's see if we can construct the 3rd column from a linear combination of the first 2.
    The third column has as its first entry:
    $$\cos2x=2\cos^2x-1$$
    That might be: ##2\cos x \times (\text{2nd column}) - (\text{1st column})##
    I checked and found it also works for the 2nd row, but the 3rd row is not so easy to check.


    So let's go to the Euler formula approach.
    Your determinant is equal to:

    $$\frac 1 8 \left| \begin{matrix}
    2 & \exp(ix) + \exp(-ix) & \exp(2ix) + \exp(-2ix) \\
    \exp(ix) + \exp(-ix) & \exp(2ix) + \exp(-2ix) & \exp(3ix) + \exp(-3ix) \\
    \exp(2ix) + \exp(-2ix) & \exp(3ix) + \exp(-3ix) & \exp(4ix) + \exp(-4ix)
    \end{matrix} \right|$$

    So let's multiply the 2nd column by ##2\cos x = \exp(ix) + \exp(-ix)##.

    $$(\exp(ix) + \exp(-ix)) \cdot \begin{pmatrix}
    \exp(ix) + \exp(-ix) \\
    \exp(2ix) + \exp(-2ix) \\
    \exp(3ix) + \exp(-3ix)
    \end{pmatrix} =
    \begin{pmatrix}
    \exp(2ix) + \exp(-2ix) & + & 2 \\
    \exp(3ix) + \exp(-3ix) & + & \exp(ix) + \exp(-ix)\\
    \exp(4ix) + \exp(-4ix) & + & \exp(2ix) + \exp(-2ix)
    \end{pmatrix}$$

    Yes... that does look somewhat like a combination of the 1st and 3rd column...

    In other words, the 3 columns are dependent, meaning the determinant is zero ##\Box##.
     
  10. Jan 20, 2013 #9

    LCKurtz

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    Of course, once you do that for any single row or column, you are done as far as proving the determinant is zero, no?
     
  11. Jan 20, 2013 #10

    I like Serena

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    Yes.
    If any single row or column is a linear combination of the others, the determinant is zero.
     
  12. Jan 20, 2013 #11

    Hurkyl

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    We can salvage Ray's argument if we can find a real solution for the y's amongst the two-dimensional space of complex solutions.

    Writing [itex]e^{ix} = a + bi[/itex], the one complex equation becomes

    [tex]y_1 + (a+ bi) y_2 + (a^2 -b ^2 + 2abi) y_3 = 0[/tex]

    And now we can read off a real solution by inspection: if we set [itex]y_3 = 1[/itex] and [itex]y_2 = -2a[/itex], then the imaginary parts cancel, leaving a real value for [itex]y_1[/itex].

    Since these y's are real, they remain a solution to the original system of equations after taking the real part.

    Now that we know the solution, we can hide away the use of complex numbers so we can look clever: the solution is
    • [itex]y_1 = -1[/itex]
    • [itex]y_2 = -2 \cos(x) [/itex]
    • [itex]y_3 = 1[/itex]
    and then we claim an identity
    [tex]-\cos nx - 2 \cos x \cos (n+1) x + \cos (n+2)x = 0[/tex]
    or to be extra fancy:
    [tex]2 \cos x \cos nx = \cos (n-1)x + \cos(n+1) x[/tex]
    which is easy to check by applying the angle addition formula on the right hand side. (Or applying the sum-of-cosines identity)
     
    Last edited: Jan 20, 2013
  13. Jan 20, 2013 #12

    Ray Vickson

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    Another way (correct this time!) is to attack the linear system Ay = 0 directly. The first equation gives ##y_1 = -\cos(x) y_2 - \cos(2x)y_3##. Substituting that into the other two equations and doing elementary trigonometric simplification gives
    [tex] (2): \; \cos(x)( -\cos(x) y_2 - \cos(2x)y_3) + \cos(2x) y_2 + \cos(3x)y_3 = 0 \Longrightarrow -\sin^2(x) (y_2 + 2 \cos(x) y_3) = 0 \\
    (3):\; \cos(2x)(-\cos(x) y_2 - \cos(2x)y_3) + \cos(3x)y_2 + \cos(4x)y_3 = 0 \Longrightarrow
    -2 \cos(x) \sin^2(x) (y_2 + 2\cos(x) y_3) = 0,[/tex]
    so for ##x \neq 0, \pi/2, \pi## we have ##y_2 = -2 \cos(x) y_3## with ##y_3## arbitrary. That means that for ##x \neq 0, \pi/2, \pi## there are non-zero solutions ##(y_1,y_2,y_3)##, so the determinant must vanish.
     
  14. Jan 20, 2013 #13
    Damn, you're right. Something else I didn't think about. Seems my linear algebra is rather rusty.
     
  15. Jan 20, 2013 #14
    That should do it. Thanks both Ray and Serena, greatly appreciated (all others who participated in the thread as well).
     
  16. Jan 20, 2013 #15

    I like Serena

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    Thank you for bringing a nice and interesting problem to our attention!
     
  17. Jan 20, 2013 #16

    Dick

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    Yours and Hurkyl's solutions are nice. They also show that the determinant of an nxn (n>=3) matrix following the same pattern is also zero.
     
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