So, on a class of matrix-analysis, we've been told that we do not define the supremum and infinum of operators. The 'problem' was the following: We have two matrices, like $$\begin{bmatrix} 1 & 0 \\ 0 & 3\\ \end{bmatrix} $$ and $$\begin{bmatrix} 2 & 0 \\ 0 & 1\\ \end{bmatrix} $$ In this problem we cannot say one is greater / smaller than the other, though we could set an infinum, eg. the identity matrix. However, as he stated, we do not define it for any operator. My question is - why? How can this be proved / did I misunderstand anything?
Supremum / infimum of operators?
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linear-algebra
matrices
supremum-and-infimum
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0In fact, one *can* (and does) define the sup/inf of a collection of positive (definite) operators. However, one needs to be careful with the ordering on the positive operators. I have not seen any ordering that applies to the space of *all* operators. – 2017-02-13
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0No ordering on the set of all operators exists since already in dimension two the space of all operators contains an isomorphic copy of the complex numbers, where order has no meaning. The positive operators correspond (in a sense) to the real numbers in the complex plane, and so it is not very surprising to find them being order-able again. – 2017-02-13
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0@IttayWeiss The order on the positive operators is just a partial ordering, though – 2017-02-13