The following results support the analogy between numbers and transformations more than anything so far; they assert that the properties that caused us to define the special classes of transformations we have been considering are reflected by their spectra.
Theorem 1. If \(A\) is a self-adjoint transformation on an inner product space, then every proper value of \(A\) is real; if \(A\) is positive, or strictly positive, then every proper value of \(A\) is positive, or strictly positive, respectively.
Proof. We may ignore the fact that the first assertion is trivial in the real case; the same proof serves to establish both assertions in both the real and the complex case. Indeed, if \(A x=\lambda x\) , with \(x \neq 0\) , then, \[\frac{(A x, x)}{\|x\|^{2}}=\frac{\lambda(x, x)}{\|x\|^{2}}=\lambda;\] it follows that if \((A x, x)\) is real (see Section: Polarization , Theorem 4), then so is \(\lambda\) , and if \((A x, x)\) is positive (or strictly positive) then so is \(\lambda\) . ◻
Theorem 2. Every root of the characteristic equation of a self-adjoint transformation on a finite-dimensional inner product space is real.
Proof. In the complex case roots of the characteristic equation are the same thing as proper values, and the result follows from Theorem 1. If \(A\) is a symmetric transformation on a Euclidean space, then its complexification \(A^{+}\) is Hermitian, and the result follows from the fact that \(A\) and \(A^{+}\) have the same characteristic equation. ◻
We observe that it is an immediate consequence of Theorem 2 that a self-adjoint transformation on a finite-dimensional inner product space always has a proper value.
Theorem 3. Every proper value of an isometry has absolute value one.
Proof. If \(U\) is an isometry, and if \(U x=\lambda x\) , with \(x \neq 0\) , then \(\|x\|=\|U x\|=|\lambda| \cdot\|x\|\) . ◻
Theorem 4. If \(A\) is either self-adjoint or isometric, then proper vectors of \(A\) belonging to distinct proper values are orthogonal.
Proof. Suppose \(A x_{1}=\lambda_{1} x_{1}\) , \(A x_{2}=\lambda_{2} x_{2}\) , \(\lambda_{1} \neq \lambda_{2}\) . If \(A\) is self-adjoint, then \[\lambda_{1}(x_{1}, x_{2})=(A x_{1}, x_{2})=(x_{1}, A x_{2})=\lambda_{2}(x_{1}, x_{2}). \tag{1}\] (The middle step makes use of the self-adjoint character of \(A\) , and the last step of the reality of \(\lambda_{2}\) .) In case \(A\) is an isometry, (1) is replaced by \[(x_{1}, x_{2})=(A x_{1}, A x_{2})=(\lambda_{1} / \lambda_{2})(x_{1}, x_{2}); \tag{2}\] recall that \(\bar{\lambda}_{2}=1 / \lambda_{2}\) . In either case \((x_{1}, x_{2}) \neq 0\) would imply that \(\lambda_{1}\) \(=\lambda_{2}\) , so that we must have \((x_{1}, x_{2})=0\) . ◻
Theorem 5. If a subspace \(\mathcal{M}\) is invariant under an isometry \(U\) on a finite-dimensional inner product space, then so is \(\mathcal{M}^{\perp}\) .
Proof. Considered on the finite-dimensional subspace \(\mathcal{M}\) , the transformation \(U\) is still an isometry, and, consequently, it is invertible. It follows that every \(x\) in \(\mathcal{M}\) may be written in the form \(x=U y\) with \(y\) in \(\mathcal{M}\) ; in other words, if \(x\) is in \(\mathcal{M}\) and if \(y=U^{-1} x\) , then \(y\) is in \(\mathcal{M}\) . Hence \(\mathcal{M}\) is invariant under \(U^{-1}=U^{*}\) . It follows from Section: Adjoints of projections , Theorem 2, that \(\mathcal{M}^{\perp}\) is invariant under \((U^{*})^{*}=U\) . ◻
We observe that the same result for self-adjoint transformations (even in not necessarily finite-dimensional spaces) is trivial, since if \(\mathcal{M}\) is invariant under \(A\) , then \(\mathcal{M}^{\perp}\) is invariant under \(A^{*}=A\) .
Theorem 6. If \(A\) is a self-adjoint transformation on a finite-dimensional inner product space, then the algebraic multiplicity of each proper value \(\lambda_{0}\) of \(A\) is equal to its geometric multiplicity, that is, to the dimension of the subspace \(\mathcal{M}\) of all solutions of \(A x=\lambda_{0} x\) .
Proof. It is clear that \(\mathcal{M}\) is invariant under \(A\) , and therefore so is \(\mathcal{M}^{\perp}\) ; let us denote by \(B\) and \(C\) the linear transformation \(A\) considered only on \(\mathcal{M}\) and \(\mathcal{M}^{\perp}\) respectively. We have \[\operatorname{det}(A-\lambda)=\operatorname{det}(B-\lambda) \cdot \operatorname{det}(C-\lambda)\] for all \(\lambda\) . Since \(B\) is a self-adjoint transformation on a finite-dimensional space, with only one proper value, namely, \(\lambda_{0}\) , it follows that \(\lambda_{0}\) must occur as a proper value of \(B\) with algebraic multiplicity equal to the dimension of \(\mathcal{M}\) . If that dimension is \(m\) , then \(\operatorname{det}(B-\lambda)=(\lambda_{0}-\lambda)^{m}\) . Since, on the other hand, \(\lambda_{0}\) is not a proper value of \(C\) at all, and since, consequently, \(\operatorname{det}(C-\lambda_{0}) \neq 0\) , we see that \(\det(A-\lambda)\) contains \((\lambda_{0}-\lambda)\) as a factor exactly \(m\) times, as was to be proved. ◻
What made this proof work was the invariance of \(\mathcal{M}^{\perp}\) and the fact that every root of the characteristic equation of \(A\) is a proper value of \(A\) . The latter assertion is true for every linear transformation on a unitary space; the following result is a consequence of these observations and of Theorem 5.
Theorem 7. If \(U\) is a unitary transformation on a finite-dimensional unitary space, then the algebraic multiplicity of each proper value of \(U\) is equal to its geometric multiplicity.
EXERCISES
Exercise 1. Give an example of a linear transformation with two non-orthogonal proper vectors belonging to distinct proper values.
Exercise 2. Give an example of a non-positive linear transformation (on a finite-dimensional unitary space) all of whose proper values are positive.
Exercise 3.
- If \(A\) is self-adjoint, then \(\operatorname{det} A\) is real.
- If \(A\) is unitary, then \(|\operatorname{det} A|=1\) .
- What can be said about the determinant of a partial isometry?