Finite-Dimensional Vector Spaces

When is a complex number $\zeta$ positive (that is, $\geq 0$ )? Two equally natural necessary and sufficient conditions are that $\zeta$ may be written in the form $\zeta=\xi^{2}$ with some real $\xi$ , or that $\zeta$ may be written in the form $\zeta=\bar{\sigma} \sigma$ with some $\sigma$ (in general complex). Remembering also the fact that (at least for unitary spaces) the Hermitian character of a transformation $A$ can be described in terms of the inner products $(A x, x)$ , we may consider any one of the three conditions below and attempt to use it as the definition of positiveness for transformations: \begin{align} &\text{$A=B^{2}$ for some self-adjoint $B$}, \tag{1}\\ &\text{$A=C^{*} C$ for some $C$}, \tag{2}\\ &\text{$A$ is self-adjoint and $(A x, x) \geq 0$ for all $x$}. \tag{3} \end{align}

Before deciding which one of these three conditions to use as definition, we observe that (1) $\implies$ (2) $\implies$ (3). Indeed: if $A=B^{2}$ and $B=B^{*}$ , then $A=B B=B^{*} B$ , and if $A=C^{*} C$ , then $A^{*}=C^{*} C=A$ and \[(A x, x)=(C^{*} C x, x)=(C x, C x)=\|C x\|^{2} \geq 0.\] It is actually true that (3) implies (1), so that the three conditions are equivalent, but we shall not be able to prove this until later. We adopt as our definition the third condition.

Definition 1. A linear transformation $A$ on an inner product space is positive , in symbols $A \geq 0$ , if it is self-adjoint and if $(A x, x) \geq 0$ for all $x$ .

More generally, we shall write $A \geq B$ (or $B \leq A$ ) whenever $A-B \geq 0$ . Although, of course, it is quite possible that the difference of two transformations that are not even self-adjoint turns out to be positive, we shall generally write inequalities for self-adjoint transformations only. Observe that for a complex inner product space a part of the definition of positiveness is superfluous; if $(A x, x) \geq 0$ for all $x$ , then, in particular, $(A x, x)$ is real for all $x$ , and, by Theorem 4 of the preceding section, $A$ must be positive.

Positive transformations are usually called non-negative semidefinite . If $A \geq 0$ and $(A x, x)=0$ implies that $x=0$ , we shall say that $A$ is strictly positive; the usual term is positive definite . Since the Schwarz inequality implies that \[|(A x, x)| \leq\|A x\| \cdot\|x\|,\] we see that if $A$ is a strictly positive transformation and if $A x=0$ , then $x=0$ , so that, on a finite-dimensional inner product space, a strictly positive transformation is invertible. We shall see later that the converse is true; if $A \geq 0$ and $A$ is invertible, then $A$ is strictly positive. It is sometimes convenient to indicate the fact that a transformation $A$ is strictly positive by writing $A>0$ ; if $A-B>0$ , we may also write $A>B$ (or \(B).

It is possible to give a matricial characterization of positive transformations; we shall postpone this discussion till later. In the meantime we shall have occasion to refer to positive matrices, meaning thereby Hermitian symmetric matrices $(\alpha_{i j})$ (that is, $\alpha_{i j}=\overline{\alpha_{j i}}$ ) with the property that for every sequence $(\xi_{1}, \ldots, \xi_{n})$ of $n$ scalars we have $\sum_{i} \sum_{j} \alpha_{ij} \xi_{i} \xi_{j} \geq 0$ . (In the real case the bars may be omitted; in the complex case Hermitian symmetry follows from the other condition.) These conditions are clearly equivalent to the condition that $(\alpha_{i j})$ be the matrix, with respect to some orthonormal coordinate system, of a positive transformation.

The algebraic rules for combining positive transformations are similar to those for self-adjoint transformations as far as sums, scalar multiples, and inverses are concerned; even Section: Self-adjoint transformations , Theorem 2, remains valid if we replace "self-adjoint" by "positive" throughout. It is also true that if $A$ and $B$ are positive, then a necessary and sufficient condition that $A B$ (or $B A$ ) be positive is that $A B=B A$ (that is, that $A$ and $B$ commute), but we shall have to postpone the proof of this statement for a while.

EXERCISES

Exercise 1. Under what conditions on a linear transformation $A$ does the function of two variables, whose value at $x$ and $y$ is $(A x, y)$ , satisfy the conditions on an inner product?

Exercise 2. Which of the following matrices are positive?

$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ .
$\begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}$ .
$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ .
$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ .
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ .

Exercise 3. For which values of $\alpha$ is the matrix \[\begin{bmatrix} \alpha & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}\] positive?

Exercise 4.

If $A$ is self-adjoint, then $\operatorname{tr} A$ is real.
If $A \geq 0$ , then $\operatorname{tr} A \geq 0$ .

Exercise 5.

Give an example of a positive matrix some of whose entries are negative.
Give an example of a non-positive matrix all of whose entries are positive.

Exercise 6. A necessary and sufficient condition that a two-by-two matrix $\big[\begin{smallmatrix}\alpha & \beta \\ \gamma & \delta\end{smallmatrix}\big]$ (considered as a linear transformation on $\mathbb{C}^{2}$ ) be positive is that it be Hermitian symmetric (that is, that $\alpha$ and $\delta$ be real and $\gamma=\bar{\beta}$ ) and that $\alpha \geq 0$ , $\delta \geq 0$ , and $\alpha \delta-\beta \gamma \geq 0$ .

Exercise 7. Associated with each sequence $(x_{1}, \ldots, x_{k})$ of $k$ vectors in an inner product space there is a $k$ -by- $k$ matrix (not a linear transformation) called the Gramian of $(x_{1}, \ldots, x_{k})$ and denoted by $G(x_{1}, \ldots, x_{k})$ ; the element in the $i$ -th row and $j$ -th column of $G(x_{1}, \ldots, x_{k})$ is the inner product $(x_{i}, x_{j})$ . Prove that every Gramian is a positive matrix.

Exercise 8. If $x$ and $y$ are non-zero vectors (in a finite-dimensional inner product space), then a necessary and sufficient condition that there exist a positive transformation $A$ such that $A x=y$ is that $(x, y)>0$ .

Exercise 9.

If the matrices $A=\big[\begin{smallmatrix}1 & 0 \\ 0 & 0\end{smallmatrix}\big]$ and $B=\big[\begin{smallmatrix}0 & 0 \\ 0 & 1\end{smallmatrix}\big]$ are considered as linear transformations on $\mathbb{C}^{2}$ , and if $C$ is a Hermitian matrix (linear transformation on $\mathbb{C}^{2}$ ) such that $A \leq C$ and $B \leq C$ , then \[C=\begin{bmatrix} 1+\epsilon & \theta \\ \theta & 1+\delta \end{bmatrix},\] where $\epsilon$ and $\delta$ are positive real numbers and $|\bar{\theta}|^{2} \leq \min \{\epsilon(1+\delta), \delta(1+\epsilon)\}$ .
If, moreover, $C \leq 1$ , then $\epsilon=\delta=\theta=0$ . In modern terminology these facts together show that Hermitian matrices with the ordering induced by the notion of positiveness do not form a lattice . In the real case, if the matrix $\big[\begin{smallmatrix}\alpha & \beta \\ \beta & \gamma\end{smallmatrix}\big]$ is interpreted as the point $(\alpha, \beta, \gamma)$ in three-dimensional space, the ordering and its non-lattice character take on an amusing geometric aspect.