Re: [線代]證方矩陣可表對稱矩陣和反對稱矩陣之和

看板Math作者Minkowski (四維之祖)時間12年前 (2012/02/08 01:31)推噓2(2推 0噓 2→)

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※ 引述《playmypig (玩我豬)》之銘言： : 證任何方形的矩皆可表為對稱矩陣和反對稱矩陣之和,而且是唯一的. : Show that for any square matrix A, there exist a symmetric matrix B and : a skew-symmetric matrix C such that A=B+C and show this expression is : ununique. : 卡關了數天也解不出來,有勞各位了,謝謝. T T T A + A A - A T A + A Let B = ──── and C = ────. Then B = ──── = B, 2 2 2 T T T T A - A A + A + A - A C = ──── = -C, and B + C = ──────── = A. 2 2 Now, to prove the uniqeuness, let A = D + E where D is symmetric T T T and E is skew-symmetric. Then A = D + E = D - E. Hence T T 2D = (D + E) + (D - E) = A + A , 2E = (D + E) - (D - E) = A - A , D = B, E = C. 　　　　　　　□ ─ 標題 [新聞] 人民幣爆炸灼傷三歲童

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