Two proofs complex matrices have eigenvalues

Today I will briefly discuss two proofs that every matrix $T$ 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 $\lambda$ such that $T - \lambda I$ 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 [1]; the second is the standard proof.

Proof by linear dependence

Let $p(x) = a_nx^n + \dots a_1x + a_0$ be a polynomial with complex coefficients. If $T$ is a linear map, $p(T) = a_nT^n + \dots a_n T + a_0I$ . We think of this as “ $p$ evaluated at $T$ ”.

Exercise: Show $(pq)(T) = p(T)q(T)$ .

Proof: Pick a random vector $v \in V$ . Consider the sequence of vectors $v, Tv, T^2v, \dots T^nv.$ This is a set of $n+1$ vectors, so they must be linearly dependent. Thus there exist constants $a_0, \dots a_n \in \C$ such that $a_nT^nv + a_{n-1}T^{n-1}v + \dots a_1Tv + a_0v$ $= (a_nT^n + a_{n-1}T^{n-1} + \dots a_1T + a_0I)v = 0$ .

Define $p(x) = a_nx^n + \dots a_1x + a_0$ . Then, we can factor $p(x) = a_n(x-\lambda_1)\dots (x-\lambda_n).$ By the Exercise, this implies $a_n(T - \lambda_1I)\dots (T-\lambda_nI)v = 0$ . So, at least one of the maps $T - \lambda_iI$ has a nontrivial kernel, so $T$ has an eigenvalue. $\square$

Proof by the characteristic polynomial

Proof: We want to show that there exists some $\lambda$ such that $T - \lambda I$ has nontrivial kernel: in other words, that $T - \lambda I$ is singular. A matrix is singular if and only if its determinant is nonzero. So, let $\chi_T(x) = \det(T - xI)$ ; this is a polynomial in $x$ , called the characteristic polynomial of $T$ . Now, every polynomial has a complex root, say $\lambda$ . This implies $T - \lambda I$ , so $T$ has an eigenvalue. $\square$

Thoughts

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!” [3].

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.

References

Sheldon Axler, Linear algebra done right. Springer 2017
Evan Chen, An Infinitely Large Napkin, available online
Sheldon Axler. Down with Determinants! The American Mathematical Monthly, 102(2), 139, 1995. doi:10.2307/2975348, available online

Arithmetic variety

Two proofs complex matrices have eigenvalues

Proof by linear dependence

Proof by the characteristic polynomial

Thoughts

References

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