The discussion in the preceding section indicates one of our reasons for wanting to study complex vector spaces. By the so-called fundamental theorem of algebra, a polynomial equation over the field of complex numbers always has at least one root; it follows that a linear transformation on a complex vector space always has at least one proper value. There are other fields, besides the field of complex numbers, over which every polynomial equation is solvable; they are called algebraically closed fields. The most general result of the kind we are after at the moment is that every linear transformation on a finite-dimensional vector space over an algebraically closed field has at least one proper value. Throughout the rest of this chapter (in the next four sections) we shall assume that our field of scalars is algebraically closed. The use we shall make of this assumption is the one just mentioned, namely, that from it we may conclude that proper values always exist.

The algebraic point of view on proper values suggests another possible definition of multiplicity. Suppose that \(A\) is a linear transformation on a finite-dimensional vector space, and suppose that \(\lambda\) is a proper value of \(A\) . We might wish to consider the multiplicity of \(\lambda\) as a root of the characteristic equation of \(A\) . This is a useful concept, which we shall call the algebraic multiplicity of \(\lambda\) , to distinguish it from our earlier, geometric , notion of multiplicity.

The two concepts of multiplicity do not coincide, as the following example shows. If \(D\) is differentiation on the space \(\mathcal{P}_{n}\) of all polynomials of degree \(\leq n-1\) , then a necessary and sufficient condition that a vector \(x\) in \(\mathcal{P}_n\) be a proper vector of \(D\) is that \(\frac{d x}{d t} \equiv \lambda x(t)\) for some complex number \(\lambda\) . We borrow from the elementary theory of differential equations the fact that every solution of this equation is a constant multiple of \(e^{\lambda t}\) . Since, unless \(\lambda=0\) , only the zero multiple of \(e^{\lambda t}\) is a polynomial (which it must be if it is to belong to \(\mathcal{P}_n\) ), we must have \(\lambda=0\) and \(x(t)=1\) . In other words, this particular transformation has only one proper value (which must therefore occur with algebraic multiplicity \(n\) ), namely, \(\lambda=0\) ; but, and this is more disturbing, the dimension of the linear manifold of solutions is exactly one. Hence if \(n>1\) , the two definitions of multiplicity give different values. (In this argument we used the simple fact that a polynomial equation of degree \(n\) over an algebraically closed field has exactly \(n\) roots, if multiplicities are suitably counted. It follows that a linear transformation on an \(n\) -dimensional vector space over such a field has exactly \(n\) proper values, counting algebraic multiplicities.)

It is quite easy to see that the geometric multiplicity of \(\lambda\) is never greater than its algebraic multiplicity. Indeed, if \(A\) is any linear transformation, if \(\lambda_{0}\) is any of its proper values, and if \(\mathcal{M}\) is the subspace of solutions of \(A x=\lambda_{0} x\) , then it is clear that \(\mathcal{M}\) is invariant under \(A\) . If \(A_{0}\) is the linear transformation \(A\) considered on \(\mathcal{M}\) only, then it is clear that \(\operatorname{det}(A_{0}-\lambda)\) is a factor of \(\det(A-\lambda)\) . If the dimension of \(\mathcal{M}\) ( \(=\) the geometric multiplicity of \(\lambda_{0}\) ) is \(m\) , then \(\operatorname{det}(A_{0}-\lambda)=(\lambda_{0}-\lambda)^{m}\) ; the desired result follows from the definition of algebraic multiplicity. It follows also that if \(\lambda_{1}, \ldots, \lambda_{p}\) are the distinct proper values of \(A\) , with respective geometric multiplicities \(m_{1}, \ldots, m_{p}\) , and if it happens that \(\sum_{i=1}^{p} m_{i}=n\) , then \(m_{i}\) is equal to the algebraic multiplicity of \(\lambda_{i}\) for each \(i=1, \ldots, p\) .

By means of proper values and their algebraic multiplicities we can characterize two interesting functions of linear transformations; one of them is the determinant and the other is something new. (Warning: these characterizations are valid only under our current assumption that the scalar field is algebraically closed.)

Let \(A\) be any linear transformation on an \(n\) -dimensional vector space, and let \(\lambda_{1}, \ldots, \lambda_{p}\) be its distinct proper values. Let us denote by \(m_{j}\) the algebraic multiplicity of \(\lambda_{j}\) , \(j=1, \ldots, p\) , so that \(m_{1}+\cdots+m_{p}=n\) . For any polynomial equation \[\alpha_{0}+\alpha_{1} \lambda+\cdots+\alpha_{n} \lambda^{n}=0,\] the product of the roots is \((-1)^{n} \alpha_{0} / \alpha_{n}\) and the sum of the roots is \(-\alpha_{n-1} / \alpha_{n}\) . Since the leading coefficient ( \(=\alpha_{n}\) ) of the characteristic polynomial \(\operatorname{det}(A-\lambda)\) is \((-1)^{n}\) and since the constant term ( \(=\alpha_{0}\) ) is \(\det(A-0)=\operatorname{det} A\) , we have \[\operatorname{det} A=\prod_{j=1}^{p} \lambda_{j}^{m_{j}}.\] This characterization of the determinant motivates the definition \[\operatorname{tr} A=\sum_{j=1}^{p} m_{j} \lambda_{j}.\] the function so defined is called the trace of \(A\) . We shall have no occasion to use trace in the sequel; we leave the derivation of the basic properties of the trace to the interested reader.

EXERCISES