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Linear Algebra & Matrix Methods
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01 — Article
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Quiz
Quiz: Vectors, Matrices, and Linear Maps
Module 1 of 5 · Easy
Quick Quiz
1.
The rank-nullity theorem states that for
A
∈
R
m
×
n
A\in\mathbb{R}^{m\times n}
A
∈
R
m
×
n
,
rank
(
A
)
+
nullity
(
A
)
\operatorname{rank}(A)+\operatorname{nullity}(A)
rank
(
A
)
+
nullity
(
A
)
equals which quantity?
n
n
n
— the number of columns of
A
A
A
.
m
m
m
— the number of rows of
A
A
A
.
m
+
n
m+n
m
+
n
— the total of rows and columns.
min
(
m
,
n
)
\min(m,n)
min
(
m
,
n
)
— the smaller dimension.
2.
Let
A
∈
R
3
×
3
A\in\mathbb{R}^{3\times3}
A
∈
R
3
×
3
have rows
(
1
,
2
,
3
)
(1,2,3)
(
1
,
2
,
3
)
,
(
4
,
5
,
6
)
(4,5,6)
(
4
,
5
,
6
)
,
(
7
,
8
,
9
)
(7,8,9)
(
7
,
8
,
9
)
. What are
rank
(
A
)
\operatorname{rank}(A)
rank
(
A
)
and
nullity
(
A
)
\operatorname{nullity}(A)
nullity
(
A
)
?
rank
=
2
,
nullity
=
0
\operatorname{rank}=2,\ \operatorname{nullity}=0
rank
=
2
,
nullity
=
0
rank
=
2
,
nullity
=
1
\operatorname{rank}=2,\ \operatorname{nullity}=1
rank
=
2
,
nullity
=
1
rank
=
1
,
nullity
=
2
\operatorname{rank}=1,\ \operatorname{nullity}=2
rank
=
1
,
nullity
=
2
rank
=
3
,
nullity
=
0
\operatorname{rank}=3,\ \operatorname{nullity}=0
rank
=
3
,
nullity
=
0
3.
The system
A
x
=
b
Ax=b
A
x
=
b
with
A
∈
R
m
×
n
A\in\mathbb{R}^{m\times n}
A
∈
R
m
×
n
has a unique solution if and only if which conditions hold?
A
A
A
is symmetric and
b
∈
R
n
b\in\mathbb{R}^n
b
∈
R
n
.
ker
(
A
)
=
{
0
}
\ker(A)=\{0\}
ker
(
A
)
=
{
0
}
and
m
≥
n
m\ge n
m
≥
n
.
b
∈
im
(
A
)
b\in\operatorname{im}(A)
b
∈
im
(
A
)
and
ker
(
A
)
=
{
0
}
\ker(A)=\{0\}
ker
(
A
)
=
{
0
}
.
b
∈
im
(
A
)
b\in\operatorname{im}(A)
b
∈
im
(
A
)
and
m
=
n
m=n
m
=
n
.
4.
A sample covariance matrix
Σ
=
1
T
X
⊤
X
\Sigma=\tfrac1T X^\top X
Σ
=
T
1
X
⊤
X
(with
X
∈
R
T
×
n
X\in\mathbb{R}^{T\times n}
X
∈
R
T
×
n
, mean-zero columns) is always:
Symmetric and positive definite.
Symmetric but not necessarily semi-definite.
Positive definite but not necessarily symmetric.
Symmetric and positive semi-definite.
5.
For
A
∈
R
5
×
8
A\in\mathbb{R}^{5\times8}
A
∈
R
5
×
8
with
rank
(
A
)
=
4
\operatorname{rank}(A)=4
rank
(
A
)
=
4
, what is the dimension of the left null space
ker
(
A
⊤
)
\ker(A^\top)
ker
(
A
⊤
)
?
$5$
$4$
$8$
$1$
6.
You bootstrap an interest-rate curve with 6 instruments and 6 unknown discount factors, and the solver returns infinitely many solutions. What is the correct diagnosis?
The discount factors must be negative, which the solver disallows.
The
6
×
6
6\times6
6
×
6
matrix
A
A
A
is rank-deficient — one instrument is redundant.
The system is overdetermined — there are more equations than there are unknowns to solve for.
A
A
A
is full rank but
b
b
b
lies outside
im
(
A
)
\operatorname{im}(A)
im
(
A
)
.
7.
To generate correlated standard-normal samples with covariance
Σ
\Sigma
Σ
(SPD), which factorisation is the correct and numerically preferred approach?
Cholesky
Σ
=
L
L
⊤
\Sigma=LL^\top
Σ
=
L
L
⊤
, then set
z
=
L
ε
z=L\varepsilon
z
=
L
ε
with
ε
∼
N
(
0
,
I
)
\varepsilon\sim\mathcal{N}(0,I)
ε
∼
N
(
0
,
I
)
.
LU with partial pivoting, then apply the
L
L
L
factor to
ε
∼
N
(
0
,
I
)
\varepsilon\sim\mathcal{N}(0,I)
ε
∼
N
(
0
,
I
)
.
Eigendecomposition
Σ
=
Q
Λ
Q
⊤
\Sigma=Q\Lambda Q^\top
Σ
=
Q
Λ
Q
⊤
, then multiply
ε
∼
N
(
0
,
I
)
\varepsilon\sim\mathcal{N}(0,I)
ε
∼
N
(
0
,
I
)
by
Q
Q
Q
alone.
Invert
Σ
\Sigma
Σ
and sample from
N
(
0
,
Σ
−
1
)
\mathcal{N}(0,\Sigma^{-1})
N
(
0
,
Σ
−
1
)
instead.
8.
A risk model uses portfolio variance
w
⊤
Σ
w
w^\top\Sigma w
w
⊤
Σ
w
where
Σ
∈
R
100
×
100
\Sigma\in\mathbb{R}^{100\times100}
Σ
∈
R
100
×
100
is estimated from 40 daily returns. Can portfolio variance be exactly zero for a non-zero portfolio?
Yes — but only if the weights sum to zero (dollar-neutral).
No —
w
⊤
Σ
w
=
0
w^\top\Sigma w=0
w
⊤
Σ
w
=
0
only when
w
=
0
w=0
w
=
0
by the definition of an inner product.
No — covariance matrices are always positive definite, so
w
⊤
Σ
w
>
0
w^\top\Sigma w>0
w
⊤
Σ
w
>
0
for all
w
≠
0
w\ne0
w
=
0
.
Yes — with
T
=
40
<
n
=
100
T=40<n=100
T
=
40
<
n
=
100
,
rank
(
Σ
)
≤
40
\operatorname{rank}(\Sigma)\le40
rank
(
Σ
)
≤
40
, so non-zero
w
∈
ker
(
Σ
)
w\in\ker(\Sigma)
w
∈
ker
(
Σ
)
give
w
⊤
Σ
w
=
0
w^\top\Sigma w=0
w
⊤
Σ
w
=
0
.
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