We have already seen ( Section: Transformations of rank one , Theorem 2) that every linear transformation A of rank \rho is the sum of \rho linear transformations of rank one. It is easy to see (using the spectral theorem) that if A is self-adjoint, or positive, then the summands may also be taken self-adjoint, or positive, respectively. We know ( Section: Transformations of rank one , Theorem 1) what the matrix of transformation of rank one has to be; what more can we say if the transformation is self-adjoint or positive?
Theorem 1. If A has rank one and is self-adjoint (or positive), then in every orthonormal coordinate system the matrix (\alpha_{i j}) of A is given by \alpha_{i j}=\kappa \beta_{i} \bar{\beta}_{j} with a real \kappa (or by \alpha_{i j}=\gamma_{i} \bar{\gamma}_{j} ). If, conversely, [A] has this form in some orthonormal coordinate system, then A has rank one and is self-adjoint (or positive).
Proof. We know that the matrix (\alpha_{i j}) of a transformation A of rank one, in any orthonormal coordinate system \mathcal{X}=\{x_{1}, \ldots, x_{n}\} , is given by \alpha_{i j}=\beta_{i} \gamma_{j} . If A is self-adjoint, we must also have \alpha_{i j}=\bar{\alpha}_{j i} , whence \beta_{i} \gamma_{j}=\overline{\beta_{j} \gamma_{i}} . If \beta_{i}=0 and \gamma_{i} \neq 0 for some i , then \bar{\beta}_{j}=\beta_{i} \gamma_{j} / \bar{\gamma}_{i}=0 for all j , whence A=0 . Since we assumed that the rank of A is one (and not zero), this is impossible. Similarly \beta_{i} \neq 0 and \gamma_{i}=0 is impossible; that is, we can find an i for which \beta_{i} \gamma_{i} \neq 0 . Using this i , we have \bar{\beta}_{j}=(\beta_{i} / \bar{\gamma}_{i}) \gamma_{j}=\kappa \gamma_{j} with some non-zero constant \kappa , independent of j . Since the diagonal elements \alpha_{j j}=(A x_{j}, x_{j})=\beta_{j} \gamma_{j} of a self-adjoint matrix are real, we can even conclude that \alpha_{i j}=\kappa \beta_{i} \bar{\beta}_{j} with a real \kappa .
If, moreover, A is positive, then we even know that \kappa \beta_{j} \bar{\beta}_{j}=\alpha_{j j}=(A x_{j}, x_{j}) is positive, and therefore so is \kappa . In this case we write \lambda=\sqrt{\kappa} ; the relation \kappa \beta_{i} \bar{\beta}_{j}=(\lambda \beta_{i})(\lambda \bar{\beta}_{j}) shows that \alpha_{i j} is given by \alpha_{i j}=\gamma_{i} \bar{\gamma}_{j} .
It is easy to see that these necessary conditions are also sufficient. If \alpha_{i j}=\kappa \beta_{i} \bar{\beta}_{j} with a real \kappa , then A is self-adjoint. If \alpha_{i j}=\gamma_{i} \bar{\gamma}_{j} , and x=\sum_{i} \xi_{i} x_{i} , then \begin{align} (A x, x) &= \sum_{i} \sum_{j} \alpha_{i j} \bar{\xi}_{i} \xi_{j}\\ &= \sum_{i} \sum_{j} \gamma_{i} \bar{\gamma}_{j} \bar{\xi}_{i} \xi_{j} \\ &= \Big(\sum_{i} \gamma_{i} \bar{\xi}_{i}\Big) \overline{\Big(\sum_{j} \gamma_{j} \bar{\xi}_{j}\Big)}\\ &= |\sum_{i} \gamma_{i} \bar{\xi}_{i}|^{2}\\ &\geq 0, \end{align}so that A is positive. ◻
As a consequence of Theorem 1 it is very easy to prove a remarkable theorem on positive matrices.
Theorem 2. If A and B are positive linear transformations whose matrices in some orthonormal coordinate system are (\alpha_{i j}) and (\beta_{i j}) respectively, then the linear transformation C , whose matrix (\gamma_{i j}) in the same coordinate system is given by \gamma_{i j}=\alpha_{i j} \beta_{i j} for all i and j , is also positive.
Proof. Since we may write both A and B as sums of positive transformations of rank one, so that \alpha_{i j}=\sum_{p} \alpha_{i}^{p} \bar{\alpha}_{j}^{p} and \beta_{i j}=\sum_{q} \beta_{i}^{q} \bar{\beta}_{j}^{q}, it follows that \gamma_{i j}=\sum_{p} \sum_{q} \alpha_{i}^{p} \beta_{i}^{q} \overline{(\alpha_{j}^{p} \beta_{j}^{q})}. (The superscripts here are not exponents.) Since a sum of positive matrices is positive, it will be sufficient to prove that, for each fixed p and q , the matrix ((\alpha_{i}^{p} \beta_{i}^{q}) \overline{(\alpha_{j}^{p} \beta_{j}^{q})}) is positive, and this follows from Theorem 1. ◻
The proof shows, by the way, that Theorem 2 remains valid if we replace "positive" by "self-adjoint" in both hypothesis and conclusion; in most applications, however, it is only the actually stated version that is useful. The matrix (\gamma_{i j}) described in Theorem 2 is called the Hadamard product of (\alpha_{i j}) and (\beta_{i j}) .
EXERCISES
Exercise 1. Suppose that \mathcal{U} and \mathcal{V} are finite-dimensional inner product spaces (both real or both complex).
- There is a unique inner product on the vector space of all bilinear forms on \mathcal{U} \oplus \mathcal{V} such that if w_{1}(x, y)=(x, x_{1})(y, y_{1}) and w_{2}(x, y)=(x, x_{2})(y, y_{2}) , then (w_{1}, w_{2})=(x_{2}, x_{1})(y_{2}, y_{1}) .
- There is a unique inner product on the tensor product \mathcal{U} \otimes \mathcal{V} such that if z_{1}=x_{1} \otimes y_{1} and z_{2}=x_{2} \otimes y_{2} , then (z_{1}, z_{2})=(x_{1}, x_{2})(y_{1}, y_{2}) .
- If \{x_{i}\} and \{y_{p}\} are orthonormal bases in \mathcal{U} and \mathcal{V} , respectively, then the vectors x_{i} \otimes y_{p} form an orthonormal basis in \mathcal{U} \otimes \mathcal{V} .
Exercise 2. Is the tensor product of two Hermitian transformations necessarily Hermitian? What about unitary transformations? What about normal transformations?
