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導き出された固有ベクトル
fushigichanの回答
ginkgoさん、こんにちは。 ＞It is readily checked that eigenvectors corresponding to λ=1 and λ=.92 are multiples of v1=[3] and v2=[ 1] [5] [-1] respectively. λ=1とλ=.92に対応する固有ベクトルがそれぞれの倍数であることを確かめるのは容易だ。 固有値λ=1に対する固有ベクトルが V1=[3] [5] 固有値λ=92に対する固有ベクトルが v2=[1] [-1] である、というのは一体どこから出てきたのか？ということですね。 固有ベクトルとは、そもそも Ax=λx となるようなベクトルxですから ＃１の方の回答どおりですが、 (A-λI)(x)=0 を解いてxを求めればいいことになります。 A=[.95 .03] [.05 .97] でしたので、 λ=1のとき、 [.95-1 .03][x1] [0] [.05 .97-1][y1] = [0] これからは -0.05x1+0.03y1=0 0.05x1-0.03y1=0 という式が導かれるので、そのような[x1,y1]の組み合わせとして [x1,y1]=[3,5]←横に書いていますが、縦だと考えてください。 を一つとってくれば充分である。 v2=[x2,y2]も同様に導き出してください。 http://www007.upp.so-net.ne.jp/masema/eigenvalue.html http://next1.cc.it-hiroshima.ac.jp/numeanal2/node10.html
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K上の有限次元線形空間Ｖに対して、写像f:V→Vは線形写像。 Ｖの基底εに対する、fの表現行列をAと表すとき、 (1)det(A) (2)Tr(A) (3)rank(A) が基底εのとり方によらず定まるということを線形変換fの固有の量 であるとのことなのですが、いまいち固有の量というものがイメージ できません固有の量とはなんなのでしょうか。 そして、なぜとり方にやらないのでしょうか？
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あら、簡単じゃないですか。←それが分からなかった奴(^^ゞ 回答を見てやっと気付きました。 (A-λI)(x)=0の式は今まで嫌というほど使ってきたのですが、 ここでこの式を使うとはこの説明からは読み取れませんでした。 ちょっと飛び過ぎの気がします、なにせ読者にとっては初めてのことを説明しているのですから…。 どうせなら「…固定ベクトルを求めてみると、それぞれの倍数であることを確かめるのは容易だ」とか書いてほしかったです>出版社さん すっきりしました。 ありがとうございました!