I remember, with a faint feeling of dread, having to calculate the eigenvalues of a matrix. It became routine in the end, but I was recently reminded of the pain when a student asked if there was a shortcut.

For a 2-by-2 matrix? Yes. It is up to you, though, whether you think it’s worth remembering.

The traditional way to find the eigenvalues of a 2-by-2 matrix is to subtract $\lambda \bf{I}$ from your matrix and find the values of $\lambda$ that make the determinant zero. You might even say that that’s the definition.

So: $\left|\mattwotwo {a &}{b}{c &}{d} - \mattwotwo {\lambda &}{0}{0 &}{\lambda}\right| = 0$, and you have to solve $(a-\lambda)(d-\lambda) - bc = 0$.

That’s simple enough to expand: $\lambda^2 - (a+d)\lambda + (ad-bc) = 0$.

You stick that in the quadratic formula, factorise, whatever it is you do to solve a quadratic, and it’s done.

### A shortcut

You could - given that this is a pattern that shows up pretty much every time you see a 2-by-2 matrix - remember the solution for it.

If you complete the square, you get $\br{\lambda - \frac{a+d}{2}}^2 - \frac{(a+d)^2}{4} + (ad-bc)=0$.

That becomes $\br{\lambda - \frac{a+d}{2}} = -ad +bc + \frac{a^2 + 2ad + d^2}{4}$.

So what? I’ll tell you what, there’s a pair of $ad$s in there that can sort their act out, that’s what. Moving the outer $-ad$ into the fraction gives:

$\br{\lambda - \frac{a+d}{2}} = bc + \frac{a^2 - 2ad + d^2}{4}$ (note the change of sign), which in turn becomes: $\br{\lambda - \frac{a+d}{2}} = bc + \frac{(a - d)^2}{4}$.

The eigenvalues are therefore $\lambda = \frac{1}{2}\br{(a+d) \pm \sqrt{4bc + (a-d)^2}}$.

### For example

Everyone’s favourite example of a 2-by-2 matrix is $\mattwotwo{1&}{2}{3&}{4}$. Its eigenvalues according to this are $\frac{1}{2}\br{ 5 \pm \sqrt{24 + 9}}$, which - happily - agrees with Wolfram|Alpha’s verdict.