[線代] rank跟subspace

[mqazz1 (無法顯示)]

98交大資訊
                         [1  1  1]
                         [1  1 -1]
suppose Q = [q1 q2 q3] = [1 -1  1]. Let S_12 = span(q1,q2)
                         [1 -1 -1]      S_23 = span(q2,q3)

(a) The union of the two subspaces S_12 and S_23 forms a vector space
        False

(c) The span(q1) is an orthogonal complement of the subspace S_23
        False

=============================

98交大資訊

Assume A is an m*n matrix with rank r and b is a column vector

(a) If m>r and n=r, then Ax=b must have no solution for some b and exactly
one solution for other b
        True

(b) If m>r and n>r, then Ax=b has infinitely many solutions for some b and
exactly one solution for other b
        False

(c) If n=r, then Ax=b has either one solution or none
        True

================================

97台大資工
                                        7*7
The set {rank(the adjoint of A) | A屬於R   } contains _____ integers.




請問這些題目要怎麼解呢?

另外像第二題(98交大的) 這種rank跟解的個數 網路上有專門介紹的嗎?

還是有哪本書有提到比較多的? 因為我是用黃子嘉考研究所的書 好像比較少提到這一塊

謝謝

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◆ From: 61.228.27.201
推 jacky7987   :黃子家的有寫啊 4-96 4-97
請問j大用的是哪一版的? 我手上是第三版的 他的4-5節矩陣的秩我有看過@@

推 jacky7987   :a)找兩個向量加出去之後不封閉的
→ jacky7987   :c)q_2和q_2不互相垂直
推 jacky7987   :上面打錯 dim(span(q_1))=1 所以complement是dim 3
推 jacky7987   :rank那幾題
→ jacky7987   :a) rank(A)=n所以是1-1 所以在CS(A)的都有解
→ jacky7987   :在N(A^T)都沒解
→ jacky7987   :因為m>r所以N(A^t)不為零空間
→ jacky7987   :b)要嘛有無限多組解要嘛無解
※ 編輯: mqazz1          來自: 61.228.27.201        (09/17 22:03)
推 jacky7987   :我是四版的耶
→ jacky7987   :不過其實他也沒介紹特別多
→ jacky7987   :但我想應該都蠻夠用的
→ jacky7987   :一個定理(算是整理過後的結論)
→ jacky7987   :=mAx=b至少一解 for all b in F^{m*1} iff rank(A)
→ jacky7987   :Ax=b至少一解 for all b in F^{m*1} iff rank(A)=m
→ jacky7987   :Ax=b至多一解 for all b in F^{m*1} iff rank(A)=n
→ jacky7987   :這個定理的上面有個注意可以看
→ jacky7987   :最後一題讓我想到我解不出來的一題
→ jacky7987   :In M_{n*n}(F), rank(A)=n then rank(adj(A))=n
→ jacky7987   :rank(A)=n-1, rank(adj(A))=1
→ jacky7987   :rank(A)<n-2, rank(adj(A))=0
→ jacky7987   :所以大概是3個整數吧(笑
推 jacky7987   :看看有沒有大大可以幫忙解一下上面那題