Adding a constant times the unit matrix and eigenvalues

[julian]

To find the eigenvalues \lambda of a matrix A you solve the equation

det |A - \lambda I| = 0 eq(1)

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

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

which is the same as solving

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

Doesn't comparison of eq(3) with eq(1) just imply \lambda_{new} = \lambda + e?

 

[conquest]

look at it this way (A+eI)v=λ_{new}v for an eigenvector v of A+eI.

Now suppose w is an eigenvector of A.

then (A+eI)w=λw+ew=(λ+e)w

so in fact even the eigenvectors coincide.

 

[julian]

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.