Self-adjoint transformations of rank one

We have already seen ( Section: Transformations of rank one , Theorem 2) that every linear transformation of rank is the sum of linear transformations of rank one. It is easy to see (using the spectral theorem) that if 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 has rank one and is self-adjoint (or positive), then in every orthonormal coordinate system the matrix of is given by with a real (or by ). If, conversely, has this form in some orthonormal coordinate system, then has rank one and is self-adjoint (or positive).

Proof. We know that the matrix of a transformation of rank one, in any orthonormal coordinate system , is given by . If is self-adjoint, we must also have , whence . If and for some , then for all , whence . Since we assumed that the rank of is one (and not zero), this is impossible. Similarly and is impossible; that is, we can find an for which . Using this , we have with some non-zero constant , independent of . Since the diagonal elements of a self-adjoint matrix are real, we can even conclude that with a real .

If, moreover, is positive, then we even know that is positive, and therefore so is . In this case we write ; the relation shows that is given by .

It is easy to see that these necessary conditions are also sufficient. If with a real , then is self-adjoint. If , and , then so that is positive. ◻

As a consequence of Theorem 1 it is very easy to prove a remarkable theorem on positive matrices.

Theorem 2. If and are positive linear transformations whose matrices in some orthonormal coordinate system are and respectively, then the linear transformation , whose matrix in the same coordinate system is given by for all and , is also positive.

Proof. Since we may write both and as sums of positive transformations of rank one, so that and it follows that (The superscripts here are not exponents.) Since a sum of positive matrices is positive, it will be sufficient to prove that, for each fixed and , the matrix 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 described in Theorem 2 is called the Hadamard product of and .

EXERCISES

Exercise 1. Suppose that and 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 such that if and , then .
There is a unique inner product on the tensor product such that if and , then .
If and are orthonormal bases in and , respectively, then the vectors form an orthonormal basis in .

Exercise 2. Is the tensor product of two Hermitian transformations necessarily Hermitian? What about unitary transformations? What about normal transformations?

EXERCISES

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