In the last few weeks, I have been learning 'Linear Algebra' in 'Khan Academy' in order to enhance my mathematics foundation. The course contains abundant concepts, knowledge points and solving skills; so it is better that I should take some notes during learning.
Determinant
- 2x2 matrix
For example, we assumeAis a 2x2 matrix. So according to the formula, the determinant of 'A' is equal to :
A = [a b]
[c d]
|A| = ad - bc
- 3x3 matrix
Turning to calculate determinant of a 3x3 matrix, the process will be more complicated.
A = [a b c]
[e f g]
[h i j]
|A| = a * (fg-ij) + (-b) * (eg-hj) + c * (ef-hi)
Inverse Matrix
If matrix A multiply another matrix B and the outcome is equal to an unit matrix, so we call matrix B is the inverse matrix of matrix A.
A * inverse(A) = E
Linear Span and Dependence
Here we have a set of vector called V, and vi belongs to V only if i is a real number.
- Linear Span
c1v1+c2v2+...+civi = VR
In the equation above, ci represent constant scalars and VR is an arbitrary vector with arbitrary dimensions. If the equation is true, so we said V can be expanded to linear span n, while n is the dimension of VR.
- Linear Dependence
The method to prove ifVis linear dependent is quite similar to that of Linear Span.
c1v1+c2v2+...+civi = 0
To make this equation true, if at least one ci has an non-zero value, then V is linear dependent; but if all 0 is the only solution to this equation, that we said V is linear independent.
Basis
Basis is a minimum set of vectors that its span is equal to the subspace of R(n). Obviously, basis is linear independent; furthermore, if there is a basis of subspace U, then any arbitrary vector in U can be only represented by an unique basis combination.Following is the proof:
1.assume A is a vector of U
2.assume a combination that can represent A
A = c1v1 + c2v2 + ... + cnvn
3.assume another combination that makes the equation true
A = d1v1 + d2v2 + ... + dnvn
4.these two equations minus each other and get a new equation
0 = (c1-d1)v1 + (c2-d2)v2 + ... + (cn-dn)vn
5.us we know, a basis is linear independent;
so only all 0 can be the solution of that equation in step 4.
In conclusion, c1 = d1, c2 = d2, ..., cn = dn.
Then we can prove that there is only one basis combination can represent a vector in its span.
