→ yueayase : 老實說我覺得你們老師的typo好多... 04/02 21:27
※ 編輯: yueayase (61.227.230.159), 04/02/2015 21:36:30
※ 編輯: yueayase (61.227.230.159), 04/02/2015 21:41:56
※ 引述《gm20090121tx (疾風破曉)》之銘言：
: 剛剛小考只會寫第三題
: 但我算出有解，等於是No free Variable
: 所以應該是No linear Combination
: 但為何題目叫我express 為何是linear Combination
: 然後其他題都不會了…
: （one-to-one跟onto那一塊完全不會QQ
: http://i.imgur.com/L5IWz4L.jpg
: 感謝大大…線代好難QAQ
: -----
: Sent from JPTT on my Sony D6653.
1.Suppose T is a linear transform
Let p=[x1] q=[x2] => T(p+q) = [y1+y2] T(p) = [y1] T(q) = [y2]
[y1] [y2] [ 1 ] [ 1] [ 1]
T(p)+T(q) = [y1+y2] ≠T(p+q)
[ 2 ]
So, T is not a linear transformation.
T T
(Remark: 只要舉反例就可以了 p=[1,0] , q=[0,1]
T T T
=> T(p+q)=[1,1], T(p)=[0,1] , T(q)=[1,1] , T(p)+T(q)=[1.2]≠T(p+q))
L=[x] = [....] 這啥東東??
[y]
2 2
姑且假設是 L([x]) = [(y+1) - (y-1) ] = [ 4y]
2 2 [-12x]
[y] [(x-3) - (x+3) ]
T T
Let p = [x1,y1] , q = [x2, y2]
T T
L(ap+bq) = L([ax1+bx2,ay1+by2]) = [4(ay1+by2), -12(ax1+bx2)]
T T
= a[4y1, -12x1] + b[4y2, -12x2]
= aT(p)+bT(q)
So, L is a linear transformation.
2. 先假設A是方陣好了
A=[a1, a2, a3] , a1, a2, a3 為A的行向量
T T
要A[5,3,-9] = [2,0,1] with rank(A) = 1
先從rank(A)=1開始想可以想見A做列運算後最後會消到只剩一排是非0列(or非0行)
要達成這樣就必須是a2=k1a1, a3=k2a1 for some nonzero k1, k2
T T
又A[5,3,-9] = 5a1+3a2-9a3 = (5+3k1-9k2)a1=[2, 0, 1]
那他只要你找一個A, 不仿就指定k1=2,k2=1
T
=> a1 = [1, 0, 0.5]
=> A=[ 1 2 1 ]
[ 0 0 0 ] 即為所求
[0.5 1 0.5]
3. 前面有人解答了
T T
4. 若只要求 B[2, 3] = [2, 0]
T T
我們不仿指定 B[0, 1] = [1, 1]
=> B[2 0] = [2 1] => B = [2 1] (1/2)[ 1 0] = (1/2)[-1 2] = [-1/2 1]
[3 1] [0 1] [0 1] [-3 2] [-3 2] [-3/2 1]
這個B就符合需求
5. 不懂題目意思... line span by (6,8) 是指方向向量是(6,8)的直線??
但這樣直線又不是唯一的
6. 雖然說不要給echelon form，但我們還是可以從echelon form出發去想
3
Ax=0的解空間是plane in R 代表 nullity(A)=2 => rank(A)=1
=> echelon form 為 [1 0 0]
[0 0 0]
那你就搞成 [2 0 0] [1 0 0] [1 1 1]
[0 0 0] [1 0 0] [1 1 1] ... 都不是echelon form
7. S很容易就可看出不是one-to-one因為映射後的結果根本沒有z
T T T
所以反例很容易找: [0,0,1], [0,0,2] 都映射到[0,0]
T 2
S是否onto? Answer: Yes. Let [s, t] in R
T 3 T T
We want to check whether there is [x,y,z] in R such that S([x,y,z])= [s,t]
T T
Since S([x,y,z]) = [x+y, x-y], we have s = x+y, t=x-y
=> x=(s+t)/2,
y=(s-t)/2
We are free to assign any z in R. For example, z = 0
So, S is onto.
T 不會是onto呢? 直覺上映射後的第三個分量限制前面兩個分量的結果，
這種情況很容易導致會有前兩個方程式會互相矛盾的情形.
T T
Counterexample: [x+y,x-y,x] = [1,1,0] => x+y=1, x-y=1, x=0
這個方程組無解，所以T不是onto
T T 2
Is T one-to-one? Yes. Let [x1,y1] , [x2,y2] in R such that
T T T T
T([x1,y1]) = T([x2,y2]). We want to show that [x1, y1] = [x2, y2].
Since [x1+y1,x1-y1,x1] = [x2+y2,x2-y2,x2], we have x1=x2, x1+y1=x2+y2,
x1-y1=x2-y2.
T T
From x1=x2, we have y1=y2. => [x1,y1] = [x2,y2]
So, T is one-to-one.
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