Self-adjoint transformations of rank one

We have already seen ( Section: Transformations of rank one , Theorem 2) that every linear transformation $A$ of rank $ρ$ is the sum of $ρ$ 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 $(α_{i j})$ of $A$ is given by $α_{i j} = κ β_{i} {\bar{β}}_{j}$ with a real $κ$ (or by $α_{i j} = γ_{i} {\bar{γ}}_{j}$ ). 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 $(α_{i j})$ of a transformation $A$ of rank one, in any orthonormal coordinate system $𝒳 = {x_{1}, \dots, x_{n}}$ , is given by $α_{i j} = β_{i} γ_{j}$ . If $A$ is self-adjoint, we must also have $α_{i j} = {\bar{α}}_{j i}$ , whence $β_{i} γ_{j} = \overset{―}{β_{j} γ_{i}}$ . If $β_{i} = 0$ and $γ_{i} \neq 0$ for some $i$ , then ${\bar{β}}_{j} = β_{i} γ_{j} / {\bar{γ}}_{i} = 0$ for all $j$ , whence $A = 0$ . Since we assumed that the rank of $A$ is one (and not zero), this is impossible. Similarly $β_{i} \neq 0$ and $γ_{i} = 0$ is impossible; that is, we can find an $i$ for which $β_{i} γ_{i} \neq 0$ . Using this $i$ , we have ${\bar{β}}_{j} = (β_{i} / {\bar{γ}}_{i}) γ_{j} = κ γ_{j}$ with some non-zero constant $κ$ , independent of $j$ . Since the diagonal elements $α_{j j} = (A x_{j}, x_{j}) = β_{j} γ_{j}$ of a self-adjoint matrix are real, we can even conclude that $α_{i j} = κ β_{i} {\bar{β}}_{j}$ with a real $κ$ .

If, moreover, $A$ is positive, then we even know that $κ β_{j} {\bar{β}}_{j} = α_{j j} = (A x_{j}, x_{j})$ is positive, and therefore so is $κ$ . In this case we write $λ = \sqrt{κ}$ ; the relation $κ β_{i} {\bar{β}}_{j} = (λ β_{i}) (λ {\bar{β}}_{j})$ shows that $α_{i j}$ is given by $α_{i j} = γ_{i} {\bar{γ}}_{j}$ .

It is easy to see that these necessary conditions are also sufficient. If $α_{i j} = κ β_{i} {\bar{β}}_{j}$ with a real $κ$ , then $A$ is self-adjoint. If $α_{i j} = γ_{i} {\bar{γ}}_{j}$ , and $x = \sum_{i} ξ_{i} x_{i}$ , 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 $(α_{i j})$ and $(β_{i j})$ respectively, then the linear transformation $C$ , whose matrix $(γ_{i j})$ in the same coordinate system is given by $γ_{i j} = α_{i j} β_{i j}$ 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 $α_{i j} = \sum_{p} α_{i}^{p} {\bar{α}}_{j}^{p}$ and $β_{i j} = \sum_{q} β_{i}^{q} {\bar{β}}_{j}^{q},$ it follows that $γ_{i j} = \sum_{p} \sum_{q} α_{i}^{p} β_{i}^{q} \overset{―}{(α_{j}^{p} β_{j}^{q})} .$ (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 $((α_{i}^{p} β_{i}^{q}) \overset{―}{(α_{j}^{p} β_{j}^{q})})$ 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 $(γ_{i j})$ described in Theorem 2 is called the Hadamard product of $(α_{i j})$ and $(β_{i j})$ .

EXERCISES

Exercise 1. Suppose that $𝒰$ and $𝒱$ 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 $𝒰 \oplus 𝒱$ 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 $𝒰 \otimes 𝒱$ 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 $𝒰$ and $𝒱$ , respectively, then the vectors $x_{i} \otimes y_{p}$ form an orthonormal basis in $𝒰 \otimes 𝒱$ .

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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