MIT Linear Algebra Study Lec 1 - 12
MIT Gilbert Strang : Lecture 1
https://www.youtube.com/watch?v=ZK3O402wf1c&list=PL49CF3715CB9EF31D&index=1
form of the matrix is Ax = b
| 2 -1 | | x | = | 0 |
| -1 2 | | y | | 3 |
the point that solves both equations is x = 1, y =2
x | 2 | + y | -1 | = | 0 |
| -1 | | 2 | | 3 |
this is a linear combination of columns
each row is plane in 3 dimensional space.
find a point that solves all the equations
planes will meet at the point.
* Ax = b can be solved with every b in this case.
Do the linear combs of the columns fill 3-dimensional space?
* for this matrix, yes.
* if some of these 3 columns lie in the same place there will not fill 3-dimensional space.
( for example, if the 3rd column is the sum of the other two columns it will not work. )
* Ax =b will also not be solved with every b.
Pretend there's 9-dimensional space with 9 vectors,
If those components are completely independent Ax = b will be true.
and it will be non-singular, and invertible.
but if they're not independent, and the 9th column is the same as the 8th column,
then it will be an 8-dimensional plane inside a 9-dimensional space.
A times X is a combination of columns of A.
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MIT Gilbert Strang : Lecture 2 ( Eliminations with Matrices )
set pivot and do eliminations.
find a second pivot and do eliminations.
then it will leave U ( Upper triangle ).
if b is | 2 | after the elimination and the result will be x = 2 , y = 1, z = -2.
| 12 |
| 2 |
matrix * columns = columns
E21
* E32 * ( E21 * A) = U == (E32 * E21) * A.
* and there will be permutation matrices if needed.
* to do column operation permutation matrix should be on the right
and to do row operation permutation matrix should be on the left.
- Inverses
E -1 * E = I
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MIT Gilbert Strang : Lecture 3 ( Multiplication and Inverse Matrices )
if A-1 exists it is invertible and nonsingular.
if there's no inverse it is a singular case.
So in Ax = 0, if x != 0, it is a singular case.
based on Gauss-Jordan's method we can get an inverse matrix.
EA = I tells us E = A -1
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MIT Gilbert Strang : Lecture 4 ( Linear Algebra )
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MIT Gilbert Strang : Lecture 5 ( Transposes, Permutations, Spaces R^n )
n ! = n ( n -1 ) .... (3)(2)(1)
Symmetric matrices
* R^T R is always symmetric.
* combinations of two-column spaces give plane. ( and it's through the origin )
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MIT Gilbert Strang : Lecture 6 ( Column Space and Nullspace )
about Vector spaces and Sub-spaces, Column space of A : Solving Ax = b.
Nullspace of A.
* What is a Vector space?
Vector space requirements
v + w and c * v are in the space.
all combinations c * v + d * w are in the space.
* In R^3 plane through the origin is a sub-space.
cause every combination of column vectors in it is in that plane.
2 subspaces : P and L
* P ∪ L = all vectors in P or L or both, is not a sub-space.
( cause v + w is not in the space)
Column space of A is a Sub-space of R^4.
* Dose Ax = b have a solution for every b?
The answer is No.
* Which b's allow this system to be solved?
in Origin, Combinations of C(A) in R^4.
this system Ax=b can be always solved when the right-hand side b
is a vector in the column space.
* And in this case 3rd column space is a combination of 1st & 2nd columns so, it's no use.
Later on, 1st and 2nd will be called pivot columns and 3rd will not.
* So the column space of this matrix will be described 2-dimensional sub-space of R^4.
* Null space of A = all Solutions x to Ax = 0 in R^3.
N(A) contains | 1 | ,
c | 1 |
| -1 |
* Check that Solutions to Ax = 0 always give a Sub-space.
The solution for this does not form a sub-space.
But the solution for the N(A) does form a sub-space.
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MIT Gilbert Strang : Lecture 7 ( Solving Ax = 0: Pivot Variables, Special Solutions )
Computing the N(A)
Pivot variables / free variables
Special Solutions - rref(A) = R
Example of the solution
1st and 2nd row is independent and 3rd row is not independent.
Because it's a combination of 1st and 2nd.
rref(A) form
* rank of A = # of pivots
For convenience to find a solution start with 1 0 matrix
then the solution is
and it can be any mulitlication of this solution.
and other solution will be 0 1 matrix, which is
* Nullspace contains all the combinations of the special solutions.
And there's one special solution for every free variable.
And the number of free variables is n(row) - r(rank).
Reduced row echelon form. zeros above & below pivots
and we can notice there's I in the pivot rows/columns.
* based on the rank we can split pivot columns and free columns.
and the magic is the combination of pivot columns and free columns gives us the solution (sub-space) of the matrix.
rref form
Rx = 0
X pivot = - F X free
Pivot variables of x is - Free variables * X position of free variables.
rank for this is 2.
number of pivot columns is 2, 1 free column
solution for this is like
* if there's only one free variable we can make only one solution, like below.
∧ this is out Nullspace matrix
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MIT Gilbert Strang : Lecture 8 ( Solving Ax = b: Row Reduced Form R )
about Vector spaces and Sub-spaces, Column space of A: Solving Ax = b.
Nullspace of A.
Let's use the same example
This is the Augmented matrix for calculation.
this is rref form and condition for solvability.
about solvability condition on b.
* 1) Ax = b is solvable when b is in C(A).
To find a complete solution to Ax = b.
1) X particular: set all free variables to zero, and solve Ax = b for pivot variables.
* make free variables zero, so rows will give calculations like this.
* and if you solve it, the result will be like this.
2) Add on anything out of X nullspace.
X = Xp (one particular solution) + Xn (any solutions from Nullspace)
* If I have anything
So the solution is,
* Nullspace subspace solutions + X particular solution.
* which means solution subspaces with the offset. (shifted away from the origin)
With this result, we can know the conditions of the matrix and solutions.
m by n matrix A of rank r ( know r ≤ m && r ≤ n )
1) Full column rank means r = n : No free variables.
N(A) = { zero vector } Solution to Ax = b
X = Xp Unique Solution if it exists ( 0 or 1 solution )
2) Full row rank means r = m
* it means every row has a pivot.
* can solve Ax=b for every b, left with n-r free variables.
r = 2
3) r = m = n
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MIT Gilbert Strang : Lecture 9 ( Independence, Basis, and Dimension )
Linear independence
Spanning a space
Basis and dimension
Suppose A is m by n with m < n.
Then there are non zero solutions to Ax = 0 (more unknown then equations)
Reason : There will be free variables.
Independence
* Vector x1, x2, ---, xn are independent if no combination gives zero vector
(except the zero combination)
* Dependent case for vectors, cause the combination of 2 vectors is on the 3rd vector.
* The columns are dependent if there is something in the null space.
like out some c1 * v1 + c2 * v2 + c3 * v3 = 0
Repeat when v1, ----, vn are columns of A.
* They are independent if nullspace of A is only the 0 vector.
( rank = n (pivot columns) )
N(A) = only zero-vector ( no free variables )
* They are dependent if A c = 0 for some non-zero c . ( rank < n )
there are free variables.
Vectors v1, ---, vl span a spcae
means : The space consists of all combinations of those vectors.
* The basis for a space is a sequence of vectors v1, v2, ---, vd with 2 properties.
1. They are independent.
2, They span the space.
Example : Space is R^3
Another basis check, for example
* R^n n vectors give a basis if the n x n matrix with those columns is invertible.
* Every basis for the space has the same number of vectors.
-> it defines a dimension of the space,.
 |
| The first 2 cols are pivots and the rest 2 cols are free variables. |
* 2 = rank (A) = # of pivot columns = dimension of C(A)
* Dimension of C(A) column space is rank and the dimension of N(A) nullspace is the number of free variables.
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MIT Gilbert Strang : Lecture 10 ( The Four Fundamental Supspaces )
1) Correct error in Lect. 9
2) Four Fundamental Subspaces (for matrix A)
1) Correct error
* The Matrix from the last lecture is not independent, cause the first two columns
are the same, so it's not invertible and it means it's dependent.
|
2) Four Subspaces
* Column space C(A) in R^m
* Nullspace N(A) in R^n
* row space = all combinations of rows of A
= transpose A and get column space of A
= all combinations of columns of A^T = C(A^T)
in R ^ m
* Nullspace of A^T = N(A^T) = left nullspace of A in R ^ n
4 Subspaces
* basis?
* dimension?
* The dimension of the C(A) is the rank (r),
* The dimension of the C(A^T) is also rank (r)
* The dimension of the N(A) is m - r
* The dimension of the N(A^T) is n - r
Examples
C(R) != C (A) different column spaces but same row spaces
The basis for row space is the first r rows of R.
EA = R ( In chapter 2, R was I, then E is A^-1 )
new vector space M
All 3x3 matrices!!
Subspaces of M || all upper triangulars || all symmetric matrices
|| diagonal matrices (dim of this subspace is 3)
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MIT Gilbert Strang : Lecture 11 ( Matrix Spaces; Rank 1; Small World Graphs)
Basis for M = all 3x3's
Graphs & Networks
if we do the elimination 4th row will be all zeros cause it's rank is 3
* The space is practically the same as 9-dimensional space.
* dim of M is 9, dim of S is 6, dim of U is 6
S ∩ U = symmetric and upper triangular = diagonal 3x3's
dim (S ∩ U) = 3
S + U (combination of S, U) = any element of S + any element of U = all 3x3's
* dim( S + U ) = 9 , dim(S) = 6, dim(U) = 6
* dim(S) + dim(U) = dim(S∩U) + dim(S+U)
complete solution : y = c1*cosx + c2*sinx
* y = cosx, sinx ( Basis ) / dim (Solution space) = 2
* for next lecture, Graph = { nodes, edges }
in this case r = 1
dim C(A) = rank = dim C(A^T)
Every rank 1 matrix has A = U(some column) * V^T (some row)
* M = all 5 x 17 matrices
Subset of rank 4 matrices, if I add two rank four matrices, is the sum rank 4??
If I add two rank 1 matrices, is that a vector space? It's most likely gonna have rank 2.
Not a subspace
it is a subspace.
= nullspace of A = [ 1 1 1 1 ]
rank = 1 = r
dim N(A) = n - r
* N(A) is R3
* C(A) is somewhere in R1
* N(A^T) = only origin.
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MIT Gilbert Strang : Lecture 12 (Graphs, Networks, Incidence Matrices)
Incidence Matrices
Kirchhoff's Laws
* dim of N(A^T) = 2
* Graph = { nodes, edges }
usage flow of water, oil
Incidence Matrix
Loops correspond to linearly dependent columns.
* If the columns are independent N(A) is only the zero vector.
Let's solve Ax = 0
make nullspace
dimensions for the N(A) is 1.
To solve this we ground the node, which means set it's potential at 0.
* so the rank of the matrix is 3. r = 3
A^T y = 0 finding N(A^T)
* A^T y = 0 Kirchoff's Current Law
* Basis for N(A^T)
Example case
* 3rd vector like below is also in the N(A^T) but it's not basis cause it's not independent. ( it's the sum of the other 2 )
* dim C(A^T) is 3 = rank
** from the N(A^T) we can know that the 3rd column is not independent and
the dim of C(A^T) is 3 so the pivot columns of this are 1st 2nd and 4th.
* and those edges correspond to the independent guys, and in the graph, those three edges have no loop.
and the name for the graph without a loop is "Tree".
"Tree": no loops
dim N(A^T) = m -r
#loops = #edges - (#nodes -1)
#nodes - #edges + #loops = 1 => Euler's formular
* nodes - edges + loop = 5 - 7 + 3 = 1
+ A^T * A is always Symetric
e = Ax , y = Ce , A^T y = f
main equations of the electronics
Basic equations of applied math.
*** A^T * C * Ax = f
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