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Anonymous asked in 科學數學 · 1 decade ago

請教線代rank與rref在null space下的疑問

請問一下這一題敘述用這種方式解釋對嗎?

(2)The rank of a m*n matrix A can not be zero, the term "full rank" means the rank of A equals m.

[自解]

No!零矩陣無rank,且full rank=min(m,n),即m,n中取最小值,不一定等於m

那想請問以下兩題是對的還是錯的?又該怎麼解釋呢?我不會寫

(7)Let Rref is the reduced row echelon form of matrix A, then the column space exactly equal to C(Rref) and the same for null space N(A)=N(Rref)

(9)IF rank of A, an m*n matrix, is n, then the nullity(i.e. dimension of null space) is 0.

不好意思,自己能力有限,比較不知變通,希望大大能用比較淺顯易懂的方式解釋,謝謝!

Update:

我漏打了,抱歉

7)Let Rref is the reduced row echelon form of matrix A, then the column space C(A) exactly equal to C(Rref) and the same for null space N(A)=N(Rref)

Update 2:

請教蜉蝣老師:

(7)row運算會破壞行關係,列運算不會破壞解的空間,這部分能否麻煩您說明一下?

(9)維度定理我們學校老師還沒教,請問您能否用別的方法說明?感謝!

1 Answer

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  • 蜉蝣
    Lv 6
    1 decade ago
    Favorite Answer

    (2)The rank of a m*n matrix A can not be zero, the term "full rank" means the rank of A equals m.

    [自解]

    No!零矩陣無rank,且full rank=min(m,n),即m,n中取最小值,不一定等於m

    基本上都是正確的,但零矩陣我們一般說rank=0不會說無rank

    (任何矩陣都有rank,我想你懂)

    --------------------------------------------------

    那想請問以下兩題是對的還是錯的?又該怎麼解釋呢?我不會寫

    (7)Let Rref is the reduced row echelon form of matrix A, then the column space exactly equal to C(Rref) and the same for null space N(A)=N(Rref)

    有問題,你指的the column space是of A?(我假設你是此意) 舉一例以下為2*2的矩陣

    |1,0| Rref => |1,0|

    |1,0| |0,0|

    ----------------------------------

    但the column space是of A為,但 C(Rref)為,

    {(x)} {(x)}

    {(x)} {(0)}

    二者並不相同,主因是row運算會破壞行關係

    若改為the row space exactly equal to R(Rref),則正確

    而N(A)=N(Rref)為正確,因為列運算不會破壞解的空間

    (9)IF rank of A, an m*n matrix, is n, then the nullity(i.e. dimension of null space) is 0.

    正確:使用維度定理知n=rank(A)+dimN(A)

    因此當rank of A, an m*n matrix, is n時

    n=n+dimN(A),=> dimN(A)=0得證

    -------------------------------

    若仍有不懂,歡迎提出

    2008-10-12 22:28:11 補充:

    To 版大,我原先就是照您後來補充的回答

    不正確的舉反例即可

    2008-10-15 18:48:15 補充:

    (7)考慮

    [1 0] [1 1]

    [1 0]行可運算=>[1 1]

    前者的解為 後者的解為

    [0] [t]

    [t] [-t]

    列運算不會破壞解的空間,這概念主要是來自於符合二關係式之解

    其必為線性組合之關係式之解

    如2x+3y=0,3x-y=0,則其解(r,s)必定為a(2x+3y)+b(3x-y)=0之解

    國中的加減消去法,就是利用此原理消去某未知數

    2008-10-15 19:04:17 補充:

    因A為m*n矩陣,N(A)的向量為n個變數所構成

    若 rank of A為 n,則列運算後只會剩n個線性獨立的rows(而其它列均為0)

    此n個rows剛好構成一可逆的n*n方陣A',因列運算不破壞解,故N(A)=N(A')

    若Ax=0<=>A'x=0,因A'為可逆,=>[A']^-1*A'x=[A']^-1*0=>x=0

    因此N(A)={0向量},因此dimN(A)=0

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