二阶矩阵的 Bellman 迹不等式

ON THE R.BELLMAN’S TRACE INEQUALITY ON 2 ORDER MATRIX

Ａ、Ｂ是二阶非负定矩阵时，证明了：ｔｒ（ＡＢ）ｎ≤ｔｒ（ＡｎＢｎ）（ｎ为自然数），此结果说明Ｒ．Ｂｅｌｍａｎ猜想对二阶矩阵成立。

This paper proves that the inequality,trace(AB) n≤trace(A nB n),is held on any 2 order non negative definite matrices A and B.This result implies that R.Bellman's guess on matrices′ trace is true on 2 order matrices.