※ [本文转录自 NTU-Exam 看板 #1M8tj5lE ]
作者: NTUForest (NTUForest) 看板: NTU-Exam
标题: [试题] 104上 吕学一 线性代数 第一次期中考
时间: Sun Oct 18 18:58:10 2015
课程名称︰线性代数
课程性质︰资讯系必修
课程教师︰吕学一
开课学院：电机资讯学院
开课系所︰资讯工程学系
考试日期（年月日）︰2015/10/18
考试时限（分钟）：180
是否需发放奖励金：是
试题 :
总共十二题，每题十分，可按任何顺序答题。只能参考个人事先准备的A4单页单面大抄。
每一题都是一个可能对也可能不对的叙述。如果你觉得对，请证明它是对的，如果你觉得
不对，请证明它是错的。(第十二题除外)课堂上证过的定理，或是提过的练习题，皆可引
用。
第一题
Let W be a vector space. If V is a non-empty subset of W, then V is a subspace
of W if and only if ax + by ∈ V holds for any scalars a,b ∈ F and any vectors
x,y ∈ V.
第二题
If(U,F,‧_U) nad (V,F,‧_V) are two vectors spaces, then so is (W,F,‧_W) with
(def)
W = F(U,V)
(def)
(f +_W g)(x) = f(x) +_V g(x)
(def)
(a ‧_W f)(x) = a‧_V(f(x))
for any f,g ∈ W, x ∈ U, and a ∈ F.
第三题
If U and V are subspaces of vector space W, then U∪V is a subspace of W if
and only if U∩V = U or U∩V = V.
第四题
If U and V are subspaces of vector space W with U + V = W and U∩V = {0_W},
then for each vector z ∈ W there exists a unique pair (x,y) with x ∈ U,
y ∈ V, and z = x + y.
第五题
Let V be a subspace of vector space W. If x is a vector of W, then the subset
{x} + V
of W is a subspace of W if and only if x ∈ V.
第六题
A subset V of vector space W is a subspace of W if and only if V + V =V.
第七题
If R and S aresubsets of vector spce V with R ⊆ S and span(R) = V, then
span(s) = V.
第八题
Let S be the subset {(1_F,0_F,0_F),(1_F,1_F,0_F),(1_F,1_F,1_F)} of vector
space F^3.
‧If F = Q, then S is linearly independent.
‧If 1_F + 1_F = 0_F, then S is linearly dependent.
第九题
If R and S are subsets of vector space V, then span(R)∩span(S)⊆span(R∩S)
if and only if R⊆S or S⊆R.
第十题
If S = {x,y,z} is a linearly independent subset of a vector space over
scalar field F, then for each s ∈ span(S) there is a unique 3-tuple
(a,b,c) ∈ F such that s = ax + by + cz.
第十一题
If x,y,z are three distinct vectors of vecror space V, then {x,y,z} is linearly
in dependent if and only if {x+y,y+z,z+x} is linearly independent.
第十二题
Prove the replacement theorem:
For any finite subset S of a vector space and any linearly independent subset Q
of span(S), there exists a subset R of S\Q with |Q| + |R| = |S| and span(Q∪R)
= span(S).