MIT Challenge 18.06

# My [MIT Challenge](https://www.scotthyoung.com/blog/myprojects/mit-challenge-2/): 18.06 Liner Algebra

[Video MIT 18.06 Linear Algebra, Spring 2005](https://www.youtube.com/watch?v=ZK3O402wf1c&list=PLE7DDD91010BC51F8) 

[MIT18_06S10_Final_Exam](https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/exams/MIT18_06S10_Final_Exam.pdf) 
[Answers](https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/exams/MIT18_06S10_Final_Answers.pdf)
```
day1~4: watch video.
40: 56-4*4. 3 hours. close.
72: 100-4*5-3*2-1. 1 day. with internet.
```
1.a
```
L'A=L'LU
L'[A I]=[U L']
L'[L I]=[I L']
L=
1 0 0
2 1 0
3 3 1
U=
1 2 0 1
0 0 1 2
0 0 0 0
```
1.b
```
1 0 0 0
0 1 0 0
0 0 0 0
2 independent cols.
```

1.c
```
[-2 1 0 0]` and [-1 0 -2 1]`
```
```
Grading:
MIT answer [3 -2 0 1] is wrong. Reason:
1. [1 2 0 1] * [3 -2 0 1]` != 0
2. Calculate online: http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=null
```
1.d
```
x=
c1[-2 1 0 0]` + c2[-1 0 -2 1]` + [1 1 1 1]
```

2.a
```
C+D=6
C+2D=4
C+4D=0
```
2.b
```
Inner product is the same:
A`p = A`b
A`Az = A`b
z = [8 -2]`
C=8
D=-2
```

2.c
```
[0 0 0]
```
```
Grading: -4
y=0 >>>> C=0, D=0. i.e. z=0.
so A`b=0, b is perpendicular with A.
y = b = c[2 -3 1]
```
3.a
```
AV=B

det(V) is n i.e. vi is independent.

A=BV'
```
3.b
```
TODO
A=sum(bivi')
```
```
Grading:
A=sum ci*bi
The column space of A consists of all linear combinations of b1, · · · , bn.
```
3.c

```
V and B is invertible i.e. vi dependent and bi dependent i.e.
det(V) = n = det(B) != 0.
A'=(BV')'=VB'
```
```
Grading:
If the bi s are independent.
me: So vi s also independent.
```
4.a
```
A=
4/5 1/10
1/5 9/10

A^k*[1 0]`:
c1(λ1^k)x1 + c2(λ2^k)x2
λ1=1 λ2=17/10-λ1=0.7
x1=[1 2]`
so limit: c1x1=[1/3 2/3]`
```
4.b
```
TODO
```
```
Grading: -4
A=
3 -4
2 -3
λ1=1 v1=[2 1]
λ2=-1 v2=[1 1]
[x'(t) y'(t)] = c1*e^λ1v1 + c2*e^λ1v1
=c1 * e^t[2 1] + c2 * e^(-t)[1 1]
[x0 y0]=[1 0] so [dx/dt|t=0 dy/dt|t=0] = [3*1 2*1] = [3 2]
c1=1
c2=1
```
4.c
```
TODO
```
```
Grading: -4
If the initial conditions are a multiple of either
eigenvector (2, 1) or (1, 1), the solution
is at all times a multiple of that eigenvector.
me: Why??? What's the "solution"? What's the formular?
```

5.a
```
a.1:
TODO

a.2:
They are perpendicular.
```
```
Grading:
j-i is orthogonal to [1 1 1]
So are k − j and i − k. By symmetry the rotation takes i to j, j to k, k to i.
me: By God i rotates to j. No need to prove?.
```

5.b
```
T=
0 0 1
1 0 0
0 1 0
(T)^3 = I. Because T is a permutation matrix that
put first col to the right. 3 times later you get an identity.
λ1+λ2+λ3=0, λλλ=1
λ1=e^(2πi/3)
λ2=e^(4πi/3)
λ3=1
```

5.c
```
1[1 0 0]` + 2[0 1 0]` + 1[0 0 1]`
```
```
Grading: -4
3D --> 2D, det(P) = 0.
λ1=0 v1=[1 2 1] perpen. plane.
For one P that does the projection:
P=
 1  0  0
 0  1  0
-1 -2  0
λ2=1 v2=[1 0 -1]
λ3=1 v3=[0 1 -1]
v2, v3 in that plane.
```
6.a
```
A`A=
14 28
28 56
λ1=0
λ2=70
V1=O
V2=1/sqrt(5)[1 2]
--------------------
AA`=
 5 10 15
10 20 30
15 30 45
λ1=0 (3 cols dependent)
λ2=70
λ3=(5+20+45)-λ2=0
v2=1/sqrt(14)[1 2 3]
v1=v3=O
```
```
Grading: -3
eigenvector shouldn't be 0.
V1=O >>>> V1=1/sqrt(5)[-2 1]
------------
v1=v3=O >>>> v1=1/sqrt(5)[-2 1 0] v3=1/sqrt(10)[-3 0 1]
```

6.b
```
V1=v1=[1 2 3]`
V2=v2-V1V1`/(V1`V1) * v2=O
q1=1/sqrt(14)[1 2 3]
q2=O
```
```
Grading:
q2 fails because col2 = 2col1 using Gram-Schmidt process.
```

6.c
```
c.1:
m>n, so r<m. AA` is m by m. rank(AA`)=rank(A)=r<m, so pivot has 0 and
eigenvalue has 0. AA` is not positive definite.
c.2:
A`A is not necessary Po. De. Test:
A=
1 0
0 0
0 0
So A`A=
1 0 0
0 0 0
0 0 0
```

```
Grading:
A=[0 0]`
```
7.a
```
det(A)=0.
Reason: row1=row2, row1 can be replaced with all 0.
det=0*subdet11+0*subdet12+... = 0
```
7.b
```
b.1:
C11=(-1)^(1+1)M11=
1 1 1
1 1 0
1 0 0
b.2:
det(B)=0*C11-1*M11=1
b.3:
volume of vectors col1~col4. volume is 1.
```
```
Grading: -1
C11=...=-1 (is a number)
volume of a box in R4 with edges = rows of B
```
7.c
```
C =>>
x 1 0 0
0 0 0 1
0 0 1 0
1 1 0 0
if x=1: det(C)=0
if x!=1: det(C)=x*(-1)-1*(-1)=1-x
combined final result is:
det(C)=1-x
```

8.a
```
Prove:
B^k=M'AM * M'AM * ... = M'(A^k)M -> M'OM -> O
```
8.b
```
S'kAS=kS'AS=kΛ is diagonal, so kA is in that space.
S'(A+B)S=(S'A+S'B)S=S'AS+S'BS=Λ1 + Λ2 is diagonal, so A+B is in that space.
Hence A form a liner subspace.
```

8.c
```
c.1:
Q=
1 0 0
0 1 0
0 0 1
c.2:
SQS'=SIS'=SS'=I
```

```
Grading: -4
Q=
1 0 0
0 1 0
0 0 1
=D1+D2+D3=
1 0 0
0 0 0
0 0 0 +

0 0 0
0 1 0
0 0 0 +

0 0 0
0 0 0
0 0 1

Then SD1S', SD2S', SD3S' are all diagonalized
by S: a basis for the subspace.
```
9.a
```
A=
-1  1  0  0
-1  0  1  0
-1  0  0  1
 0 -1  1  0
 0 -1  0  1
 0  0 -1  1
rank(A)=3 by reduce.
dim(N(A`))=m-r=6-3=3
```

9.b
```
b.1:
A`A=
 3 -1 -1 -1
-1  3 -1 -1
-1 -1  3 -1
-1 -1 -1  3

b.2:
rank(A`A)=3

b.3:
nullspace: 4-3=1
```
```
Grading: -3
nullspace: c[1 1 1 1]
```
9.c
```
A`A is above.
V=c[1 1 1 1]. Wrong problem. "my apologies!"
```