Linear Algebra – Axler’s Proof of Eigenvalue Existence

abstract-algebradeterminanteigenvalues-eigenvectorslinear algebrapolynomials

I am puzzled by Sheldon Axler's proof that every linear operator on a finite dimensional complex vector space has an eigenvalue (theorem 5.10 in "Linear Algebra Done Right"). In particular, it's his maneuver in the last set of displayed equations where he substitutes the linear operator $T$ for the complex variable $z$. See below.


Theorem 5.10 in "Linear Algebra Done Right", Sheldon Axler


What principle allows Axler to make this substitution of linear operator for complex variable and still claim that the RHS equals the LHS, and that the LHS goes from being the number zero (in the preceding set of equations) to the zero vector (in the final set of equations)?

As a possible answer to my own question, on the preceding page he offers these remarks:


polynomial over linear operator


Is linearity of the map from "polynomial over complex variable" to "polynomial over T" the property that makes this work? If so, I don't feel very enlightened. Any further insight here is greatly appreciated, especially since Axler makes a side comment that this proof is much clearer than the usual proof via the determinant.

Best Answer

The essential point here is not just linearity, but also compatibility with multiplication.

This means the following. When $$p(X) = \sum_n a_n X^n, \qquad q(X) = \sum_n b_n X^n$$ we define multiplication of polynomials in the usual way by $$(pq)(X) = \sum_n c_n X^n, \quad \text{ where } c_n = \sum_{i + j = n} a_i b_j.$$

Then not only do we have $$(pq)(z) = p(z)q(z)$$ for any $z \in \mathbf{C}$, but also $$(pq)(T) = p(T)q(T)$$ for any vector space endomorphism $T$.

By induction, this can be extended to products of an arbitrary finite number of polynomials.

Thus the equality of polynomials $$a_0 + a_1 X + \ldots + a_n X^n = c(X - \lambda_1) \cdots (X - \lambda_n)$$ entails both $$a_0 + a_1 z + \ldots + a_n z^n = c(z - \lambda_1) \cdots (z - \lambda_n)$$ and $$a_0 I+ a_1 T + \ldots + a_n T^n = c(T - \lambda_1 I) \cdots (T - \lambda_n I)$$ In abstract terms, for any $\mathbf{C}$-algebra $A$ and an element $a \in A$, the evaluation map $\operatorname{ev}_a \colon \mathbf{C}[X] \to A, \ p \mapsto p(a)$ is a morphism of $\mathbf{C}$-algebras. This holds in particular both for $A = \mathbf{C}$ and for $A = L(V)$.

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