Linear Algebra Proofs
SI Leader: Mr.L. Khanyile
02 June 2025
1. In a vector space V , show that k · 0 = 0 for all scalars k.
Solution:
k · 0 = k · (0 + 0)
=k·0+k·0
⇒k·0−k·0=0
⇒k·0=0
2. Prove that the set S of all n × n matrices A such that AT = −A
is a subspace of Mn .
Solution:
Let A, B ∈ S so that AT = −A and B T = −B. Let c ∈ R. Then:
(A + B)T = AT + B T = −A − B = −(A + B)
(cA)T = cAT = −cA
So S is closed under addition and scalar multiplication. Therefore, S is a subspace Mn .
3. Prove that if S = {v1 , v2 , v3 } is linearly dependent and v4 is any
vector in V not in S, then S ′ = {v1 , v2 , v3 , v4 } is also linearly depen-
dent.
Solution:
Since S is linearly dependent, there exist scalars a1 , a2 , a3 , not all zero, such that
a1 v1 + a2 v2 + a3 v3 = 0.
Then
a1 v1 + a2 v2 + a3 v3 + 0 · v4 = 0,
with not all scalars zero, so S ′ is linearly dependent.
4. Let A be an n × n matrix that is diagonalizable. Prove that A
has n linearly independent eigenvectors.
Solution:
Since A is diagonalizable, there exists an invertible matrix P such that:
P −1 AP = D
where D is diagonal with entries λ1 , . . . , λn . Then:
AP = P D
1
Let columns of P be p1 , . . . , pn , so:
Api = λi pi for 1 ≤ i ≤ n
Since P is invertible, its columns are linearly independent. Therefore, A has n linearly independent eigen-
vectors.
5. Let T : V → W be a linear transformation. Prove that the range
of T is a subspace of W .
Solution:
The range of T is defined as Range(T ) = {T (v) : v ∈ V } ⊆ W .
(1) Zero vector: Since T is linear,
T (0V ) = 0W ∈ Range(T ).
(2) Closed under addition and scalar multiplication: Let w1 = T (v1 ) and w2 = T (v2 ) for v1 , v2 ∈ V ,
and let c ∈ R. Then
w1 + cw2 = T (v1 ) + cT (v2 ) = T (v1 + cv2 ) ∈ Range(T ).
Since the range of T contains the zero vector and is closed under addition and scalar multiplication. Thus,
it is a subspace of W .
6. In a vector space V , show that the zero element is unique.
Solution:
Suppose 0 and 0′ are both zero elements. Then for any v ∈ V :
v + 0 = v = v + 0′ ⇒ 0 = 0′
Therefore, the zero element is unique.
7. Prove that the set S = {A ∈ Mn | AB = BA} is a subspace for
fixed B ∈ Mn .
Solution:
The zero matrix satisfies 0 · B = 0 = B · 0, so 0 ∈ S.
Let A1 , A2 ∈ S and c ∈ R. Then:
(A1 + A2 )B = A1 B + A2 B = BA1 + BA2 = B(A1 + A2 ),
(cA1 )B = c(A1 B) = c(BA1 ) = B(cA1 ).
So A1 + A2 ∈ S and cA1 ∈ S, showing closure under addition and scalar multiplication.
Hence, S is a subspace of Mn .
2
8. Let A be an n × n matrix and S = {v1 , . . . , vk } a set of eigenvectors
corresponding to distinct eigenvalues λ1 , . . . , λk . Prove that S is
linearly independent.
Solution:
Assume for contradiction that {v1 , . . . , vk } is linearly dependent. Let r be the largest index such that
{v1 , . . . , vr } is linearly independent. Then {v1 , . . . , vr+1 } is dependent, so:
c1 v1 + · · · + cr+1 vr+1 = 0 (1)
Apply A to both sides:
A(c1 v1 + · · · + cr+1 vr+1 ) = A(0) = 0
A(c1 v1 + · · · + cr+1 vr+1 ) = c1 Av1 + · · · + cr+1 Avr+1 = 0
Since each vi is an eigenvector of A with eigenvalue λi , this becomes:
c1 λ1 v1 + · · · + cr+1 λr+1 vr+1 = 0 (2)
Multiply (1) by λr+1 and subtract (2):
c1 (λr+1 − λ1 )v1 + · · · + cr (λr+1 − λr )vr = 0 (3)
Since {v1 , . . . , vr } is linearly independent and all λi ̸= λr+1 , it follows that:
c1 = · · · = cr = 0
Substitute into (1):
cr+1 vr+1 = 0 ⇒ cr+1 = 0
This contradicts the assumption that not all ci are zero. Therefore, {v1 , . . . , vk } is linearly independent.
