# 33. Eigenexamples

## 33. Eigenexamples

Example:

Let A = 2, 1; 1, 1. The characteristic polynomial is det of A minus t I equals det of 2 minus t, 1; 1, 1 minus t, which equals (2 minus t) times (1 minus t), minus 1, which equals t squared minus 3 t + 1 The roots of this polynomial are 3 plus or minus the square root of 9 minus 4, all over 2, i.e. 3 plus or minus root 5 all over 2.

Rather than doing the plus and minus cases separately, I'll just leave plus or minus in the notation; at the end, the answer will end up with a plus or minus sign in and you pick the sign corresponding to the eigenvalue you want. The eigenvectors for the eigenvalue 3 plus or minus root 5, all over 2 are the solutions to A v = lambda v, i.e. 2, 1; 1, 1 times x, y equals (3 plus or minus root 5, all over 2) times x, y Multiplying out, this gives the equations 2 x + y equals (3 plus or minus root 5), all times x over 2 and x + y = (3 plus or minus root 5), all times y over 2. Rearranging, we get y = (minus 1 plus or minus root 5), all times x over 2, and x = (1 plus or minus root 5), all times y over 2 As usual, the second equation follows from the first (check it!), so our eigenvectors are those of the form x, (minus 1 plus or minus root 5), all times x over 2.

What this notation means is that the eigenvectors for (3 plus root 5), all over 2 are x, (minus 1 plus root 5), all times x over 2 and the eigenvectors for (3 minus root 5), all over 2 are x, (minus 1 minus root 5), all times x over 2. Another useful piece of notation is minus or plus (which means minus, (plus or minus)).

Example:

Let A = 3 over 2, 5 over 2, 3; minus 1 over 2, minus 3 over 2, minus 3; 1, 1, 2. The characteristic polynomial is: det of A minus t I equals det of 3 over 2 minus t, 5 over 2, 3; minus 1 over 2, minus 3 over 2 minus t, minus 3; 1, 1, 2 minus t, which equals (3 over 2 minus t) times det of minus 3 over 2 minus t, minus 3; 1, 2 minus t, minus 5 over 2 times det of minus 1 over 2, minus 3; 1, 2 minus t, plus 3 times det of minus 1 over 2, minus 3 over 2 minus t; 1, 1, which equals, after much working minus t cubed + 2 t squared + t minus 2. To find the roots of this cubic, we use a standard trick: we guess one of them! Let's guess t = 1. Amazingly this is a root (funny how often that happens in carefully designed examples) because minus 1 cubed + 2 times 1 squared + 1 minus 2 = 0. Once we have this, we know that our cubic factors as (t minus 1) times a quadratic. Now we get the quadratic by doing polynomial long division. I don't know a good way of typesetting this, so you'll have to look at the video for the working! Note the similarities between long division and row reduction! These are both special cases of a general technique called "Groebner bases".

The result is: minus t squared + t + 2 This has roots minus plus or minus square root of 9, all over minus 2, i.e. minus 1 or 2.

This means that the eigenvalues are minus 1, 1 and 2. What are the eigenvectors? I'll just work out the eigenvectors for lambda = 1 and leave the others as a fun exercise (answers at the end of the video).

We want to solve A v = v, which means 3 x over 2 + 5 y over 2 + 3 z = x, minus x over 2 minus 3 y over 2 minus 3 x = y, x + y + 2 z = z The first two rows are both equivalent to x + 5 y + 6 z = 0, the third means z = minus x minus y, so overall we get y = minus 5 x and z = 4 x In other words, the eigenvectors have the form x, minus 5 x, 4 x Answers for lambda = 2 and minus 1 from .