课程名称︰线性代数
课程性质︰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）︰104/1/11
考试时限（分钟）：180分钟
试题 :
台大资工线性代数单双合璧 期末考
2015年1月11日下午2:20起三个小时
总共十一题，每题十分，可按任何顺序答题。可参考个人事先准备的A4单页单面的大抄。
第2～11题是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。课堂上证过的定理、提过的习题、之前的考题，都可以直接引用。
第一题 Compute an orthonormal basis of span(S) with respect to the standard
inner product of M_1x4(C), where
S = {(1, i, 2 - i, -1),(2 + 3i, 3i, 1 - i, 2i),
(-1 + 7i, 6 + 10i, 11 - 4i, 3 + 4i)}.
第二题 The T : P_2(R) → P_2(R) defined by
T(f(x)) = f(0) + f(1)(x + x^2)
is a diagonalizable linear operator on P2(R).
第三题 Let β be a basis of an inner-product space V with dim(V) < ∞. Let x
and y be two vectors of V. If
〈x|z〉=〈y|z〉
holds for all vectors z ∈ β, then
x = y.
第四题 If Y is an orthonormal subset of an inner-product space V with
dim(V) < ∞, then
2 2
∥x∥ ≧ Σ |〈x|y〉|
y∈Y
holds for any vector x ∈ V. (Recall |a + bi| = √(a^2 + b^2).)
第五题 Let T be an invertible linear operator on a vector space V with
dim(V) < ∞. If λ is an eigenvalue of T, then λ^-1 is an eigenvalue of T^-1.
第六题 Let T be an invertible linear operator on a vector space V with
dim(V) < ∞. If λ is an eigenvalue of T and λ^-1 is an eigenvalue of T^-1,
then E_T(λ) = E_(T^-1)(λ^-1).
第七题 Let T be an invertible linear operator on a vector space V with
dim(V) < ∞. If T is diagonalizable, then so is T^-1.
第八题 For any two similar matrices A and B in M_nxn(F), there are
‧ a vector space V over F with dim(V) = n,
‧ a linear operator T on V , and
‧ ordered bases β and γ of V
β γ
with A = [T] and B = [T] .
β γ
第九题 If A ∈ M_nxn(F) and a_i,j with 1 ≦ i, j ≦ n is the element of A in
the i-th row and the j-column, then
n n ~
Σ sgn(σ) Π a = Σ a ．det(A ),
σ∈S_n i=1 i,σ(i) i=1 i,i i,i
~
where A_i,i is the submatrix of A obtained by deleting the i-th row and the
i-column of A.
第十题 If B is a square matrix that can be obtained by performing exactly one
elementary row operation on a square matrix A, then B can also be obtained by
performing exactly one elementary column operation on A.
第十一题 If B ∈ M_5x5(R) with
╭ 1 0 -1 2 1 ╮
│-1 1 3 -1 0 │ x B = 0 ,
│-2 1 4 -1 3 │ M_4x5(R)
╰ 3 -1 -5 1 -6 ╯
then
rank(B) ≦ 2.