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Determinant of this symmetric matrix (proof)

  1. May 11, 2005 #1
    Hello all.

    I'm stuck with this excercise that is asking me to proof that the determinant of the nxn matrix with a's on the diagonal and everywhere else 1's equals to:

    |A| = (a + n - 1).(a-1)^(n-1)

    So the matrix should look something like:
    [a 1 1.. 1]
    [1 a 1.. 1]
    [: ....... :]
    [1 ..1 1 a]

    I started subtracting row n-1 from row n, row n-2 from row n-1 and so on. But this gives me a matrix with a bidiagonal part (if that's the correct term, probably not!) under the first row. This looks needlessly complicated, and I don't know how to go from there. Maybe I did it the wrong way..

    I would appreciate any help. :smile:
     
    Last edited: May 11, 2005
  2. jcsd
  3. May 11, 2005 #2

    mathwonk

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    try induction?
     
  4. May 11, 2005 #3
    Yes, I've thought of that idd. I'll try to work it out.

    But since this excercise comes out of a high school math book on matrices, any other suggestions?

    Thanks!
     
  5. May 11, 2005 #4

    mathwonk

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    i think even high schoolers know induction. and it looks ,like the easiest way to me.

    the power of induction is you do not have to understand why it is true, you just crank it out. that makes it especially suitable for naive students.
     
  6. May 17, 2005 #5
    I was able to solve it using induction. :biggrin:
    Had to use cofactor expansion to write the n_determinant as a sum of n-1_determinants and then it gave me this nice little formula to use. :smile:

    Tnx again!

    edit: pardon my typo in the first post. :tongue2:
     
    Last edited: May 17, 2005
  7. Sep 18, 2006 #6
    diagonal and ones determinant

    =====
    Hi, I have the same task, can you save me and show me the solution. Thanks.
     
  8. Feb 4, 2010 #7
    Hello !

    I am new on the forum but I will try to help as i can.

    The case a=1 tells you in that case that the Kernel has dimension n-1 (only one free direction given by column vector with only 1 in it). So that you now that the determinant has to include (a-1)^(n-1). You just miss the last direction information but it is easily given if you notice that the column vector 1 is sent into itself provinding the factor (a+n-1)

    Another way of saying the same : diagonalise it ! to do so in a smart way : compute the square of the matrix, and you will see it can be expressed as a combination of the same matrix and the identity. This equation provides you a polynom that cancels the matrix. All the eigenvalues are roots of this polynom and the determinant is the product of them (with their actual multiplicity i.e. you will find n-1 times (a-1))
     
  9. Jun 2, 2010 #8
    Simplest way: The following linearly independent vectors are eigenvectors:

    [1,....,1]', [1 -1,0,..0], [0, 1 -1 0,...0].....[0,...0,1 -1].

    The eigenvalue corresponding tho the first is a+n-1. That corresponding to each of the remaining n-1 is a-1. Thus as determinant is the product of eigenvalues you get the result.
     
  10. Jun 4, 2010 #9
    I doubt the OP cares since this thread is 5 years old, but another simple way to do it is to add columns 2,3,...,n to column 1 and then subtract row 1 from every other row. This results in a triangular matrix.
     
    Last edited: Jun 5, 2010
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