Av=λv has a nonzero solution if and only if λ is a root of the characteristic polynomial of A : det(A-tI)=0.
32. Finding eigenvalues
32. Finding eigenvalues
Characteristic polynomial
In the last video, we introduced the equation Av=λv . For each λ , this gives us an equation for v . The question we will now answer is: for which λ∈𝐂 does this equation have a nonzero solution v ?
Here, t is just a dummy variable we've introduced (not one of the components of v or anything like that). The characteristic polynomial is a polynomial in t of degree n . We'll do some examples, then prove the theorem.
Examples
Suppose A=(2-110) . The characteristic polynomial is: det(A-tI)=det((2-110)-t(1001))
You will come to love the formula for solving quadratics; it lets you find the eigenvalues of any 2-by-2 matrix. By contrast, a 3-by-3 matrix will have a cubic characteristic polynomial. Whilst there is a formula for solving cubics, it's not nice. For 4-by-4 matrices, it gets still worse. For 5-by-5 and bigger matrices, the characteristic polynomial is a quintic or higher degree polynomial, and there's (provably) no general formula for the solution of a general quintic in terms of taking k th roots etc.
Let A=(0-110) . Then det(A-tI)=det(-t-11-t)=t2+1.
Let's figure out the eigenvectors. For λ=i , we need to solve Av=iv : (0-110)(xy)=(ixiy).
For λ=-i , we need to solve Av=-iv , which gives y=ix , and the eigenvectors are those of the form (xix) .
Proof of theorem
If there exists a nonzero solution v to Av=λv then (A-λI)v=0 . This implies that A-λI is not invertible; otherwise we get v=(A-λI)-10=0 . Therefore det(A-λI)=0 , so λ is a root of det(A-tI) .
In fact, these are all "if and only if" statements. The only nonobvious one is to see that if A-λI is not invertible then there exists a nonzero v such that (A-λI)v=0 (you might like to think about why that's true).