Re: [線代] projection matrix 一點不懂

看板Math作者cwhalf (c-w)時間10年前 (2014/04/24 19:57)推噓2(2推 0噓 10→)

留言12則, 6人參與討論串4/4 (看更多)

喔我突然懂了..... 抱歉請容我重新 formulate 一下： Vectors a1 and a2 are the basis of a plane. The plane of a1 and a2 equals to the column space of A. That is, the first column in the matrix A is a1, and the second column of it is a2. There is another vector called b. I want to know the projection (called p) of b on this plane. Of course if b is in the column space of A (i.e. b equals to the linear combinations of a1 and a2), p=b. But unfortunately b is not in this column space. 到此為止我的理解應該都沒錯吧？但目前為止這個 A 的 dimension = 2 也就是它是一個二維的平面經由一連串的公式推導可以得出 p=A*(AtA)^(-1)*At*b=P*b [這個大寫的 P 叫做 projection matrix; At = transpose of A; ^(-1) = inverse; * = multiplication] 這個時候這個 MIT 的老師說除非這種特殊情況： If A is a square and invertible matrix, then P equals to the identify matrix. Moreover, b is already in the column space of A, and the projection of b into the whole space of A is actually b itself. 這是因為當 A 是 symmetric and invertible 時他的 size 必定是 n x n 且這 n 個 columns 是 independent 的它的 dimension 就變成 n 而非原來的 2 了所以 b 當然就是 A 的一個 column space b 在 A 這個 space 的 p 就是 b 本身而 P 就是 identity matrix 囉我想應該這樣沒錯了吧？ PS. 其實我只是想要了解 eigenvalue/vector 以及 PCA 有需要懂這些 space 到這麼多嗎? Thanks a lot! -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.121.113 ※ 文章網址: http://www.ptt.cc/bbs/Math/M.1398340621.A.5CE.html

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cwhalf

04/24 20:01, , 1^F

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cwhalf

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loveann

04/24 21:14, , 3^F

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