Cub11k's BIU Notes
Cub11k's BIU Notes
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Discrete-math
Discrete-math 1
Discrete-math 10
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Infi-1
Infi-1 10
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Linear-1
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Linear-2
Linear-2 1
Lectures
Data-structures
Data-structures 1
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Discrete-math
Discrete-math 10
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Discrete-math 13
Discrete-math 14
Discrete-math 15
Discrete-math 16
Discrete-math 18
Discrete-math 19
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Discrete-math 24
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Discrete-math 3
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Discrete-math 7
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Exam 2023 (2A)
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Exam 2023 (A)
Exam 2023 (B)
Exam 2023 (C)
Exam 2024 (A)
Exam 2024 (B)
Exam 2024 (C)
Midterm
Infi-1
Exam 2022B (A)
Exam 2022B (B)
Exam 2023B (A)
Exam 2023B (B)
Exam 2024 (A)
Exam 2024 (B)
Exam 2025 (A)
Infi-1 10
Infi-1 12
Infi-1 13
Infi-1 14
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Infi-1 19
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Infi-1 23
Infi-1 24
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Infi-1 26
Infi-1 5
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Midterm
Theorems and proofs
Infi-2
Infi-2 1
Infi-2 10
Infi-2 11
Infi-2 12
Infi-2 13
Infi-2 14
Infi-2 15
Infi-2 16
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Infi-2 2-3
Infi-2 3-4
Infi-2 5
Infi-2 6
Infi-2 7
Infi-2 8
Infi-2 9
Linear-1
Exam 2023 (B)
Exam 2023 (C)
Exam 2024 (A)
Exam 2024 (B)
Exam 2024 (C)
Exam 2025 (A)
Linear-1 11
Linear-1 12
Linear-1 13
Linear-1 4
Linear-1 5
Linear-1 6
Linear-1 7
Linear-1 8
Linear-1 9
Midterm
Random exams
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Linear-2
Linear-2 1
Linear-2 2
Linear-2 3
Linear-2 4
Linear-2 5
Linear-2 6
Linear-2 7
Linear-2 8
Seminars
CSI
CSI 2
Data-structures
Data-structures 1
Data-structures 2
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Discrete-math
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Infi-1
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Infi-1 3
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Infi-2
Infi-2 1
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Linear-1
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Linear-2
Linear-2 1
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Templates
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Linear-2 2
Let
A
∈
R
n
×
n
Prove or disprove:
A
invertible
⟺
A
A
T
invertible
Proof:
det
(
A
)
≠
0
⟺
det
(
A
T
)
≠
0
⟺
det
(
A
A
T
)
≠
0
Let
A
∈
R
n
×
n
be symmetric
Prove or disprove:
A
invertible
⟺
A
+
A
T
invertible
Proof:
A
+
A
T
=
2
A
det
(
A
)
≠
0
⟺
det
(
2
A
)
=
2
n
det
(
A
)
≠
0
⟺
det
(
A
+
A
T
)
≠
0
Let
α
∈
R
Let
A
∈
R
n
×
n
A
i
j
=
{
α
i
=
j
1
i
≠
j
|
α
1
…
1
1
α
…
1
⋮
⋮
⋱
⋮
1
…
1
α
|
→
∀
i
∈
[
2
,
n
]
:
R
1
+
R
i
|
α
+
n
−
1
α
+
n
−
1
…
α
+
n
−
1
1
α
…
1
⋮
⋮
⋱
⋮
1
…
1
α
|
→
1
α
+
n
−
1
R
1
(
α
+
n
−
1
)
|
1
1
…
1
1
α
…
1
⋮
⋮
⋱
⋮
1
…
1
α
|
→
∀
i
∈
[
2
,
n
]
:
R
i
−
R
1
(
α
+
n
−
1
)
|
1
1
…
1
0
α
−
1
…
0
⋮
⋱
⋱
⋮
0
…
0
α
−
1
|
⟹
det
(
A
)
=
(
α
+
n
−
1
)
(
α
−
1
)
n
−
1
Let
A
∈
F
n
×
n
Let
λ
∈
F
λ
is called an eigenvalue of
A
⟺
∃
v
∈
F
n
≠
0
:
A
v
=
λ
A
v
is then called an eigenvector of
A
in respect to eigenvalue
λ
λ
≠
0
⟹
A
(
1
λ
v
)
=
v
⟹
v
∈
C
(
A
)
λ
=
0
⟹
A
v
=
0
⟹
v
∈
N
(
A
)
Characteristic polynomial
Characteristic polynomial
P
A
(
λ
)
=
|
λ
I
−
A
|
|
λ
−
5
6
−
3
λ
+
4
|
=
(
λ
−
5
)
(
λ
+
4
)
+
18
=
(
λ
+
1
)
(
λ
−
2
)
P
A
(
λ
)
=
0
⟺
λ
is an eigenvalue of
A
Proof:
∃
v
≠
0
:
A
v
=
λ
v
⟺
(
λ
I
−
A
)
v
=
0
⟺
v
∈
N
(
λ
I
−
A
)
⟺
det
(
λ
I
−
A
)
=
0
⟺
P
A
(
λ
)
=
0
A
=
(
0
1
−
1
0
)
P
A
(
λ
)
=
|
λ
−
1
1
λ
|
=
λ
2
+
1
A
∈
R
n
×
n
⟹
No eigenvalues
A
∈
C
n
×
n
⟹
±
i
is an eigenvalue
A
∈
Z
2
n
×
n
⟹
1
is an eigenvalue
A
=
(
3
1
1
2
4
2
1
1
3
)
|
λ
−
3
−
1
−
1
−
2
λ
−
4
−
2
−
1
−
1
λ
−
3
|
→
R
1
+
R
2
+
R
3
|
λ
−
6
λ
−
6
λ
−
6
−
2
λ
−
4
−
2
−
1
−
1
λ
−
3
|
=
(
λ
−
6
)
|
1
1
1
−
2
λ
−
4
−
2
−
1
−
1
λ
−
3
|
→
(
λ
−
6
)
|
1
1
1
0
λ
−
2
−
2
0
0
λ
−
2
|
=
(
λ
−
6
)
(
λ
−
2
)
2
P
A
(
λ
)
=
0
⟺
[
λ
=
6
λ
=
2
λ
=
6
⟹
v
λ
=
N
(
0
0
0
−
2
2
−
2
−
1
−
1
3
)
=
N
(
0
0
0
−
1
1
−
1
0
−
1
2
)
=
N
(
0
0
0
−
1
0
1
0
−
1
2
)
=
=
s
p
{
(
1
2
1
)
}
λ
=
2
⟹
v
λ
=
N
(
−
1
−
1
−
1
−
2
−
2
−
2
−
1
−
1
−
1
)
=
N
(
1
1
1
0
0
0
0
0
0
)
=
s
p
{
(
−
1
1
0
)
,
(
−
1
0
1
)
}
Let
A
∼
B
P
A
(
λ
)
=
P
B
(
λ
)
Proof:
|
λ
I
−
B
|
=
|
λ
I
−
P
−
1
A
P
|
=
|
P
−
1
(
λ
I
−
A
)
P
|
=
=
|
P
−
1
|
⋅
|
λ
I
−
A
|
⋅
|
P
|
=
|
λ
I
−
A
|
Let
T
:
V
→
V
be a linear transformation
∀
B
basis of
V
:
λ
is an eigenvalue of
T
⟺
λ
is an eigenvalue of
[
T
]
B
Proof:
Let
v
≠
0
∈
V
T
v
=
λ
v
⟺
[
T
v
]
B
=
[
λ
v
]
B
⟺
[
T
]
B
[
v
]
B
=
λ
[
v
]
B