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## Inverse of Matrix 
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So far we have been working on given $X_n, A$, find $X_{n+1}$\\
Can we do it reversely?
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e288fc6a-68ac-41a3-bced-38f3084f04f2
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For a matrix A, and vectors X and B\\
AX = B\\
If I give you B and A, can you find X for me?\\
Yes!!!!\\
X = $A^{-1}B$
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$AX = B \implies X = A^{-1}B \ne BA^{-1}$ (order matters)\\
e.g.
A=$\begin{bmatrix}1&2\\3&4\end{bmatrix}, B = \begin{bmatrix}4\\2\end{bmatrix}, X = \begin{bmatrix}x_1\\x_2\end{bmatrix}$\\
$AX = \begin{bmatrix}x_1+2x_2\\3x_1+4x_2\end{bmatrix} = \begin{bmatrix}4\\2\end{bmatrix}\implies x_1=-6,x_2 = 4$\\ (Wait, this solution has to deal with B's value, can we find a expression for all B's?)
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Question: Can we express it with a matrix? ($A^{-1}$) (So we can solve it without knowing B)\\
A=$\begin{bmatrix}1&2\\3&4\end{bmatrix}, B = \begin{bmatrix}b_1\\b_2\end{bmatrix}, X = \begin{bmatrix}x_1\\x_2\end{bmatrix}$\\
$AX = \begin{bmatrix}x_1+2x_2\\3x_1+4x_2\end{bmatrix} = \begin{bmatrix}b_1\\b_2\end{bmatrix}\implies x_1=b_2-2b_1,x_2 = \frac{3}{2}b_1-\frac{1}{2}b_2$\\
$\implies \begin{bmatrix}-2&1\\1.5&-0.5\end{bmatrix}B = X$

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︠84479bac-063d-4566-935e-ecbacc1a934c︠

A = matrix(RDF,[[1,2],[3,4]])
B = vector([4,2])
X = A.inverse()*B
X
A*X
A.inverse()
︡03887e93-7c12-4892-b45e-a1d17e4f0ce6︡{"stdout":"(-5.999999999999998, 4.999999999999999)\n"}︡{"stdout":"(4.0, 2.0000000000000036)\n"}︡{"stdout":"[-1.9999999999999996 0.9999999999999998]\n[ 1.4999999999999998 -0.4999999999999999]\n"}︡{"done":true}
︠8d991002-6463-4d78-a402-846004e16280︠
@interact
def FindX(x1=(-7,7,1),x2=(-7,7,1)):

 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)

︡8c01b01b-4123-4feb-9e7e-9ca39a5e440a︡{"interact":{"controls":[{"animate":true,"control_type":"slider","default":0,"display_value":true,"label":"x1","vals":["-7","-6","-5","-4","-3","-2","-1","0","1","2","3","4","5","6","7"],"var":"x1","width":null},{"animate":true,"control_type":"slider","default":0,"display_value":true,"label":"x2","vals":["-7","-6","-5","-4","-3","-2","-1","0","1","2","3","4","5","6","7"],"var":"x2","width":null}],"flicker":false,"id":"cc261eaa-5aed-40b2-831f-e12251127ef6","layout":[[["x1",12,null]],[["x2",12,null]],[["",12,null]]],"style":"None"}}︡{"done":true}
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$A^{-1} = \frac{1}{A}$\\
$A\times A^{-1} = I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$, like 1\\
(p.s. Inverse does not always exist)\\
For scalar, if a =0, $\frac{1}{a}$ does not exist!!\\
So when does $A^{-1}$ do not exist??
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## Change of Basis 
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Any vector we see now is actually a combination of basis vectors!!!\\
$\begin{bmatrix}x_1\\x_2\end{bmatrix} = x_1\begin{bmatrix}1\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\end{bmatrix}$\\
For a matrix $T = \begin{bmatrix}t_1&t_2\end{bmatrix}$\\
$T\begin{bmatrix}x_1\\x_2\end{bmatrix} = x_1t_1+x_2t_2$\\
T maps the original vector $X$, from the world $t_1,t_2$ to another world of $\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}$ \\
$\implies T^{-1}$ maps from world $\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}$ back to $t_1,t_2$
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## 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)

︡568b6a23-5238-4888-8157-496978673126︡{"hide":"input"}︡{"md":"## Sometimes $T^{-1}$ Does not exist\n## 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 $\\begin{bmatrix}0\\\\1\\end{bmatrix}$\n## Cannot map back!!!\n\n### $\\implies T^{-1}$ exists if and only if all $t_1, t_2, t_3......$ cannot be form by any other basis except itself.\n#### Impossible to find $t_1 = at_2 + bt_3$ for any a and b. (3 dimensional case)"}︡{"done":true}
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## Diagonalization
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︠06436a6d-7a35-433a-a437-1865ab5c674ei︠
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A is a 2x2 matrix, has eigen values $\lambda_1,\lambda_2$, eigen vectors $V_1=\begin{bmatrix}x_1\\y_1\end{bmatrix},V_2=\begin{bmatrix}x_2\\y_2\end{bmatrix}$\\
What is $A\begin{bmatrix}V_1&V_2\end{bmatrix}$?\\
Question:\\ $A\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}= $\\
(a) $\begin{bmatrix}\lambda_1&\lambda_2\\\lambda_1&\lambda_2\end{bmatrix}\begin{bmatrix}x_1\\y_1\end{bmatrix}$\\
(b) $\begin{bmatrix}\lambda_2&\lambda_1\\\lambda_1&\lambda_2\end{bmatrix}\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}$\\
(c) $\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}$\\
(d) $\begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$\\
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$T = \begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}$\\
$AT = T\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$\\
$ T^{-1}AT=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$
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## Longterm Effect 
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︠cd8d0a1c-d6cf-4951-b2c6-a9b7e968f1bci︠
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\begin{flushleft}
$X_{n+1} = AX_n$\\
$X_{n+2} = AX_{n+1} = A^2X_n$\\
$X_{n+3} = AX_{n+2} = A^3X_n$\\
$X_{n+2} = AX_{n+3} = A^4X_n$\\
$X_{n+2} = AX_{n+4} = A^5X_n$\\
$X_{n+2} = AX_{n+5} = A^6X_n$
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︠efd711cf-59a3-4ec6-b02c-216ce5b716e6i︠
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## Can we find $A^6$? 
$T^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT$
 
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︠7fc41ec3-0f25-4c9d-8976-fed860d0d4a0i︠
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## Can we find $A^6$? 
$T^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT$ = $T^{-1}A^6T$
︡e3aea645-6879-47f4-b6e0-02c2b833a517︡{"hide":"input"}︡{"md":"## Can we find $A^6$? \n$T^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}ATT^{-1}AT$ = $T^{-1}A^6T$"}︡{"done":true}
︠a9bf9046-63d2-49bf-be10-9f82d7d2b978i︠
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$T = \begin{bmatrix}x_1&x_2\\y_1&y_2\end{bmatrix}$\\
$AT = T\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$\\
$ T^{-1}AT=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$\\
$ T^{-1}A^6T=\begin{bmatrix}\lambda_1^6&0\\0&\lambda_2^6\end{bmatrix}$\\
$ A^6=T\begin{bmatrix}\lambda_1^6&0\\0&\lambda_2^6\end{bmatrix}T^{-1}$
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︠88e48eb0-ca70-49c1-9b3f-bf1e60e2be1ds︠

A= matrix(RDF,[[1,1],[2,0.5]])
b1 = vector([2,3])
b2 = vector([-5,1.2])
b3 = vector([-4,-8])

A.inverse()*b1
A.inverse()*b2
A.inverse()*b3
︡703867a6-0904-4157-8176-c62e1b57d491︡{"stdout":"(1.3333333333333335, 0.6666666666666665)\n"}︡{"stdout":"(2.4666666666666663, -7.466666666666666)\n"}︡{"stdout":"(-4.0, 0.0)\n"}︡{"done":true}
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︠b1d12cb0-5b78-4d2f-8fbe-e5ce296eaa43︠

Collaborative Calculation and Data Science

colby

sagemath

cornellcollege

facebook

Inverse of Matrix

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 $A^6$ ?

Can we find $A^6$ ?

Inverse of Matrix

Change of Basis

Sometimes T−1T^{-1}T−1 Does not exist

If T=[t1t1]T =[t_1 t_1]T=[t1​t1​] => multiple vectors in T's world maps to the same vector in world [10]\begin{bmatrix}1\\0\end{bmatrix}[10​] and [01]\begin{bmatrix}0\\1\end{bmatrix}[01​]

Cannot map back!!!

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

Impossible to find t1=at2+bt3t_1 = at_2 + bt_3t1​=at2​+bt3​ for any a and b. (3 dimensional case)

Diagonalization

Longterm Effect

Can we find A6A^6A6?

Can we find A6A^6A6?

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 $\begin{bmatrix}0\\1\end{bmatrix}$

$\implies T^{-1}$ 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)

Can we find $A^6$ ?

Can we find $A^6$ ?