To find the eigenvalues [itex]\lambda[/itex] of a matrix [itex]A[/itex] you solve the equation(adsbygoogle = window.adsbygoogle || []).push({});

[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]?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Adding a constant times the unit matrix and eigenvalues

Loading...

Similar Threads - Adding constant times | Date |
---|---|

Deriving the structure constants of the SO(n) group | Nov 19, 2015 |

Why adding a multiple of one column to another does not change the determinant of a matrix | Oct 11, 2014 |

Motivating Def./Formula for Adding, Multiplying Matrices, Without Appealing to Linear | Feb 14, 2011 |

How to prove that the dim of two subspaces added together equals dim | Mar 30, 2010 |

Matric adding up to zero | Dec 6, 2009 |

**Physics Forums - The Fusion of Science and Community**