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]?

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# Adding a constant times the unit matrix and eigenvalues

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