Today I will briefly discuss two proofs that every matrix over the complex numbers (or more generally, over an algebraically closed field) has an eigenvalue. Notice that this is equivalent to finding a complex number such that has nontrivial kernel. The first proof uses facts about “linear dependence” and the second uses determinants and the characteristic polynomial. The first proof is drawn from Axler’s textbook ; the second is the standard proof.
Proof by linear dependence
Let be a polynomial with complex coefficients. If is a linear map, . We think of this as “ evaluated at ”.
Exercise: Show .
Proof: Pick a random vector . Consider the sequence of vectors This is a set of vectors, so they must be linearly dependent. Thus there exist constants such that .
Define . Then, we can factor By the Exercise, this implies . So, at least one of the maps has a nontrivial kernel, so has an eigenvalue.
Proof by the characteristic polynomial
Proof: We want to show that there exists some such that has nontrivial kernel: in other words, that is singular. A matrix is singular if and only if its determinant is nonzero. So, let ; this is a polynomial in , called the characteristic polynomial of . Now, every polynomial has a complex root, say . This implies , so has an eigenvalue.
To me, it seems like the determinant based proof is more straightforward, although it requires more machinery. Also, the determinant based proof is “constructive”, in that we can actually find all the eigenvalues by factoring the characteristic polynomial. On subject of determinant-based vs determinant-free approaches to linear algebra, see Axler’s article “Down With Determinants!” .
There is a similar situation for the problem of showing that the sum (or product) of two algebraic numbers is algebraic. Here there is a non-constructive proof using “linear dependence” (which I attempted to describe in a previous post) and a constructive proof using the characteristic polynomial (which will hopefully be the subject of a future blog post). A further advantage of the determinant-based proof is that it can be used more generally to show that the sum and product of integral elements over a ring are integral. In this more general context, we no longer have linear dependence available.