[理工] [工數] [矩陣] 特徵值

看板Grad-ProbAsk作者iamwwj (阿Ken)時間14年前 (2010/03/23 20:45)推噓5(5推 0噓 19→)

留言24則, 4人參與討論串1/2 (看更多)

(a) Given a matrix A = [ 5 -1 ] , find the orthonormal matrix T [ -1 5 ] that can produce the diagonal matrix D = [λ1 0] by D = TAT^-1, [0 λ2] where λ1 and λ2 are eigenvalues and λ1 > λ2 <=====(3) 我的計算過程如下: 目測λ= 6 , 6 將λ= 6 代回 (A - λI) T = 0 , [ -1 -1 ][T1] = 0 [ -1 -1 ][T2] = 0 得到 T = c1 [ 1] + c2 [-1] = [ 1 -1] <========= (1) [-1] [ 1] [-1 1] D = TAT^-1 = [ 6 0 ] <========= (2) [ 0 6 ] --------------------------------------------------------------------- (1)題目中的orthonormal矩陣T,我其實不知道在問什麼, 我是解讀成求特徵向量,所以用特徵向量解法計算,請問這觀念是否正確? 請問錯誤的話orthonormal矩陣T該如何求出呢? (2)這裡對角矩陣 D 的 TAT^-1計算過程我直接跳過, ￣￣￣￣￣￣￣把求到的λ1與λ2代進對角線之中, 不按順序放入λ1與λ2, 此做法是否合宜? (λ皆為同數值情況下 or 非同數值情況下) (3)說的是如果λ非同數值,則λ1放較大的數值嗎?? ---------------------------------------------------------------------- (b) Given the quadratic form X^T A X = Q where X = [X1,X2]^T , prove that λ1 y1^2 +λ2 y2^2 = Q if y = T^-1 X ,where y =[y1,y2]^T ----------------------------------------------------------------- 我的計算過程如下: X^T = [X1,X2] 由(a)小題得知 A = [ 5 -1] [-1 5] [ 5 -1] [5 X1^2 - X2^2] X^T A X = [X1,X2] [-1 5] [X1,X2]^T = [- X1^2 +5 X2^2] <=========(4) 將(a)小題求得的T代入 y = T^-1 X T=[ 1 -1] y =[ 1 -1]^-1 X = 1/2 [1 1] X╮ [-1 1] [-1 1] [1 1] ├ y = [y1 y1] │ [y2 y2] (把題目給的y代入上式) y = [y1,y2]^T ╯ ---------- X 2 由上式可得知 y1 = 1/2 , y2 = 1/2 將(a)小題求得的λ1,λ2 = 6 代入 λ1 y1^2 +λ2 y2^2 = Q 6 x y1^2 + 6 x y2^2 = 3 ----------------------------------------------------- (4)我算出λ1 y1^2 +λ2 y2^2 的 Q = 3 了可是好像跟X^T A X 的 Q 沒相關性 ,請問我答案3就是題目要的答案嗎? 因為這是連鎖題,感覺(a)錯了後面(b)就跟著錯了,寫的很沒把握 ------------------------------------------------------- (c) Identify the conic section 5X1^2 - 2X1X2 + 5X2^2 = 24 and plot the graph the conic section. 這題好像是畫直角座標系的圖,可是不知如何下筆有勞版上朋友賜教了 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 220.139.140.118

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t5d

03/23 20:48, , 1^F

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iamwwj

03/23 20:49, , 2^F

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squallting

03/23 21:13, , 3^F

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