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Adding a constant times the unit matrix and eigenvalues

  1. Apr 7, 2012 #1

    julian

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    To find the eigenvalues [itex]\lambda[/itex] of a matrix [itex]A[/itex] you solve the equation

    [itex]det |A - \lambda I| = 0[/itex] eq(1)

    but now what if you add [itex]e I[/itex] to the matrix A where e is a constant? Then you have to solve the equation,

    [itex]det |(A + eI) - \lambda_{new} I| = 0[/itex] eq(2)

    which is the same as solving

    [itex]det |A - (\lambda_{new} - e) I| = 0[/itex] eq(3)

    Doesn't comparison of eq(3) with eq(1) just imply [itex]\lambda_{new} = \lambda + e[/itex]?
     
    Last edited: Apr 8, 2012
  2. jcsd
  3. Apr 8, 2012 #2
    look at it this way [itex](A+eI)v=λ_{new}v[/itex] for an eigenvector v of A+eI.

    Now suppose w is an eigenvector of A.

    then [itex](A+eI)w=λw+ew=(λ+e)w[/itex]

    so in fact even the eigenvectors coincide.
     
  4. Apr 8, 2012 #3

    julian

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    Thanks cus a PhD in maths didn't know what I was talking about....I was just checking I was right. Your way is more transparent.
     
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