X-Received: by 10.107.11.222 with SMTP id 91mr18767714iol.7.1454762811798; Sat, 06 Feb 2016 04:46:51 -0800 (PST) X-Received: by 10.182.153.2 with SMTP id vc2mr349388obb.11.1454762811775; Sat, 06 Feb 2016 04:46:51 -0800 (PST) Path: news.ccsf.jp!tomockey.ddo.jp!border1.nntp.dca1.giganews.com!border2.nntp.dca1.giganews.com!nntp.giganews.com!hb3no1343468igb.0!news-out.google.com!kr2ni8252igb.0!nntp.google.com!o2no2648034iga.0!postnews.google.com!glegroupsg2000goo.googlegroups.com!not-for-mail Newsgroups: fj.sci.math Date: Sat, 6 Feb 2016 04:46:51 -0800 (PST) In-Reply-To: Complaints-To: groups-abuse@google.com Injection-Info: glegroupsg2000goo.googlegroups.com; posting-host=133.16.216.36; posting-account=vqDRSAoAAAC6TG7fw5Br3gMzzlpRlKaf NNTP-Posting-Host: 133.16.216.36 References: <62922cc6-d799-464e-8273-19f57a58f445@googlegroups.com> <91dfc981-a144-4ed8-8045-785fe35ef436@googlegroups.com> <17b9e435-a3c7-42db-a3b1-9dc2e43a1ce4@googlegroups.com> <784fa791-b878-4ce1-bc3e-977e23e85f37@googlegroups.com> <43730c9c-98f2-4fb8-aecc-c9c126153823@googlegroups.com> <8d451cb0-8da7-4bae-86bb-4b060de03052@googlegroups.com> <08ce7704-e1bb-46a1-a015-4d8940085f93@googlegroups.com> <543c9bc4-9607-47b2-aff5-60887bc131a3@googlegroups.com> User-Agent: G2/1.0 MIME-Version: 1.0 Message-ID: Subject: =?ISO-2022-JP?B?UmU6IDMbJEIhXxsoQjMbJEJANUNNJSglayVfITwlSDlUTnMkTkA1Q01ALRsoQg==?= From: chiaki@kit.jp Injection-Date: Sat, 06 Feb 2016 12:46:51 +0000 Content-Type: text/plain; charset=ISO-2022-JP Content-Transfer-Encoding: 7bit Lines: 34 Xref: news.ccsf.jp fj.sci.math:3962 工繊大の塚本です. 2016年2月5日金曜日 9時09分14秒 UTC+9 Kyoko Yoshida: > 「0≠∀w∈∧^2C^3に対して,0≠∃k∈C,∃u_2,u_3∈C^3 such that > ∃u_1∈C^3;{u_1,u_2,u_3}はC^3の正規直交基底」の理解に難儀しております。 > k_1≠0の時と,k_1=0且つ(k_2,k_3)≠(0,0)の時について計算してみました > (下記の(iii)の箇所です)。 > http://www.geocities.jp/kyokoyoshi0515/questions/GrassmannProduct.pdf > 双方ともk=1となってしまったのですがこれでいいのでしょうか? k_1 \neq 0 のとき, k_1 e_1Λe_2 + k_2 e_1Λe_3 + k_3 e_2Λe_3 = (e_1 - (k_3/k_1) e_3)Λ(k_1 e_2 + k_2 e_3) であり, v_1 = e_1 - (k_3/k_1) e_3, v_2 = k_1 e_2 + k_2 e_3 とおくと, = v_1Λv_2 ですが, v_1, v_2 を正規直交化して得られる u_2 = (1/\sqrt{1 + |k_3/k_1|^2})(e_1 - (k_3/k_1) e_3), u_3 = (1/\sqrt{|k_2|^2|k_3|^2/|k_1|^2 + (1+|k_3/k_1|^2)^2|k_1|^2 + |k_2|^2}) \times (k_2\bar{k_3}/\bar{k_1} e_1 + (1+|k_3/k_1|^2)k_1 e_2 + k_2 e_3) について, v_1 = \sqrt{1 + |k_3/k_1|^2} u_2, v_2 = (- k_2\bar{k_3}/\bar{k_1}) \sqrt{1 + |k_3/k_1|^2} u_2 + \sqrt{|k_2|^2|k_3|^2/|k_1|^2 + (1 + |k_3/k_1|^2)^2|k_1|^2 + |k_2|^2} \times(1/\sqrt{1+|k_3/k_1|^2}) u_3 ですから, v_1Λv_2 = \sqrt{|k_2|^2|k_3|^2/|k_1|^2 + (1 + |k_3/k_1|^2)^2|k_1|^2 + |k_2|^2} u_2Λu_3 となり, k = \sqrt{|k_2|^2|k_3|^2/|k_1|^2 + (1 + |k_3/k_1|^2)^2|k_1|^2 + |k_2|^2} でしょう. k_1 = 0 の場合はお考え下さい. -- 塚本千秋@基盤科学系.京都工芸繊維大学 Tsukamoto, C. : chiaki@kit.ac.jp