Honors Linear Algebra-Homework 3

span(S1) ∩ span(S2) are equal and on in which they are unequal.
Proof. S1 ∩ S2 ⊂ S1 ⊂ span(S1). That means span(S1) is a subspace
containing S1 ∩ S2. So we have span(S1 ∩ S2) ⊂ span(S1). Similarly, span(S1 ∩
S2) ⊂ span(S2). Hence span(S1 ∩ S2) ⊂ span(S1) ∩ spac(S2).
We can simply take S1 = S2, then S1 ∩ S2 = S1 = S2, and span(S1 ∩ S2) =
span(S1) = span(S1) ∩ span(S2).
For an example such that span(S1 ∩ S2) ̸= span(S1) ∩ spac(S2), we can take
V = R and S1 = {1} and S2 = {2}. Then S1 ∩ S2 = ∅, so span(S1 ∩ S2) = {0}.
But span(S1) = span(S2) = R, hence span(S1) ∩ span(S2) = R.
2 1.5 Linear Dependence and Linear Independence
3.In M3×2(F ), prove that the set
⎧⎛
⎨ 1
⎝0 ⎩
0
⎞ ⎛
1 0
0⎠
, ⎝1 0 0
⎞ ⎛
0 0
1⎠
, ⎝0 0 1
⎞ ⎛
0 1
0⎠
, ⎝1 1 1
⎞ ⎛
0 0
0⎠
, ⎝0 0 0
⎞⎫
1 ⎬
1⎠
⎭
1
is linearly dependent.
Proof. We have
⎛
1
⎝0 ⎞ ⎛
1 0
0⎠
+ ⎝1 ⎞ ⎛
0 0
1⎠
+ ⎝0 ⎞ ⎛
0 1
0⎠
− ⎝1 ⎞ ⎛
0 0
0⎠
− ⎝0 ⎞ ⎛
1 0
1⎠
= ⎝0 ⎞
0
0⎠
.
0 0 0 0 1 1 1 0 0 1 0 0
Hence they are linearly dependent.
6.In Mm×n(F ), let E ij denote the matrix whose only nonzero entry is 1 in
the ith row and jth column. Prove that {E ij : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is linearly
independent.
Proof.Suppose there exists {aij|aij ∈ F 1 ≤ i ≤ m, 1 ≤ j ≤ n} such that
m∑ n∑
aijE ij = 0. Left hand side is a matrix whose entry in ith row and jth
i=1 j=1
column is aij. Right hand side is the zero matrix. They are equal if and only if
aij = 0, ∀1 ≤ i ≤ m, 1 ≤ j ≤ n. Hence {E ij : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is linearly
independent.
8.Let S = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} be a subset of the vector space F 3 .
(a)Prove that if F = R, then S is linearly independent.
(b)Prove that if F has characteristic 2, then S is linearly dependent.
2