[理工] [工數] [矩陣] 特徵值
(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 ]
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(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??
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(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
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(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
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(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
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