Lab06_slides2.sagews

Inverse of Matrix

A = matrix(RDF,[[1,2],[3,4]])
B = vector([4,2])
X = A.inverse()*B
X
A*X
A.inverse()

(-5.999999999999998, 4.999999999999999) (4.0, 2.0000000000000036) [-1.9999999999999996 0.9999999999999998] [ 1.4999999999999998 -0.4999999999999999]

@interact
def FindX(target = vector([4,2]),x1=(-7,7,1),x2=(-7,7,1)):
    B = vector([4,2])
    v1 = vector([1,3])
    v2 = vector([2,4])
    vres = v1*x1+v2*x2;
    fig = plot(v1*x1,color='red',linestyle='--')+plot(v2*x2,color='blue',linestyle='--')+plot(vres,color='green')+point(B,size=30,color='black')
    show(fig)

Interact: please open in CoCalc

Change of Basis

Sometimes $T^{-1}$ Does not exist

If $T =[t_1 t_1]$ => multiple vectors in T's world maps to the same vector in world $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$

Cannot map back!!!

$\implies T^{-1}$ exists if and only if all $t_1, t_2, t_3......$ cannot be form by any other basis except itself.

Impossible to find $t_1 = at_2 + bt_3$ for any a and b. (3 dimensional case)

Diagonalization

Longterm Effect

Can we find $A^6$?

$T^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT$

Can we find $A^6$?

$T^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT$ = $T^{-1}A^6T$

T = matrix(RDF,[[1,1],[2,0.5]])
v1 = vector([2,3])
T.inverse()*v1
v2 = vector([-5,1.2])
v3 = vector([-4,-8])
T.inverse()*v2
T.inverse()*v3

(1.3333333333333335, 0.6666666666666665) (2.4666666666666663, -7.466666666666666) (-4.0, 0.0)

@interact 
def findR()