SharedTA Sandbox / Hao's Sandbox / Lab06_slides / Lab06_slides2.sagewsOpen in CoCalc
Author: HAO LEE

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 T1T^{-1} Does not exist

If T=[t1t1]T =[t_1 t_1] => multiple vectors in T's world maps to the same vector in world [10]\begin{bmatrix}1\\0\end{bmatrix} and [01]\begin{bmatrix}0\\1\end{bmatrix}

Cannot map back!!!

    T1\implies T^{-1} exists if and only if all t1,t2,t3......t_1, t_2, t_3...... cannot be form by any other basis except itself.

Impossible to find t1=at2+bt3t_1 = at_2 + bt_3 for any a and b. (3 dimensional case)

Diagonalization

Longterm Effect

Can we find A6A^6?

T1ATT1ATT1ATT1ATT1ATT1ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT














Can we find A6A^6?

T1ATT1ATT1ATT1ATT1ATT1ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT = T1A6TT^{-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()