- #1
julian
Gold Member
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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]?
[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]?
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