Why are eigenvalues offset by a constant of 0.5?

Question

If you use the quadratic formula to find a general equation for eigenvalues you get this:

$ M= \left[ {\begin{array}{cc} 1 & b \\ c & d \\ \end{array} } \right] $

$v =$ eigenvalue(s) of the matrix (quadratic)

So:

$det(M)=0$
$(a-v)(d-v)-(b)(c)=0$
$v^2-av-dv+ad-bc=0$
$(1)v^2+(-a-d)v+(ad-bc)=0$

So if we use the quadratic formula:

$h=1$
$j=-a-d$
$k=ad-bc$
$v=\frac{-j\pm\sqrt{j^2-4hk}}{2h}$

Expanding we get:

$v=\frac{a+d\pm\sqrt{a^2+2ad+d^2-4ad+4bc}}{2}$

Then symplifying...

$v=\frac{1}{2}+\frac{d\pm\sqrt{a^2+d^2+4bc-2ad}}{2}$

My question is: is there an intuitive reason the value is offset by a constant 0.5? Or is this just one of those "it just is" things?

Answer 1

The product of the eigenvalues equals the determinant, and the sum of the eigenvalues equals the trace. The trace of $M$ is $1+d$, which means the average of the two eigenvalues is $\frac{1+d}{2}$. Your explicit calculation backs that up.