# Adding a constant times the unit matrix and eigenvalues

1. Apr 7, 2012

### 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 $\lambda_{new} = \lambda + e$?

Last edited: Apr 8, 2012
2. Apr 8, 2012

### 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.

3. Apr 8, 2012

### 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.