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from sympy import *
init_printing()
Use row_del and row_insert to go from one Matrix to the other.
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def matrix1(M):
"""
>>> M = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
>>> M
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
>>> matrix1(M)
[4, 5, 6]
[0, 0, 0]
[7, 8, 9]
"""
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M = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
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M
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matrix1(M)
Use the matrix constructors to construct the following matrices. There may be more than one correct answer.
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def matrix2():
"""
>>> matrix2()
[4, 0, 0]
[0, 4, 0]
[0, 0, 4]
"""
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matrix2()
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def matrix3():
"""
>>> matrix3()
[1, 1, 1, 0]
[1, 1, 1, 0]
[0, 0, 0, 1]
"""
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matrix3()
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def matrix4():
"""
>>> matrix4()
[-1, -1, -1, 0, 0, 0]
[-1, -1, -1, 0, 0, 0]
[-1, -1, -1, 0, 0, 0]
[ 0, 0, 0, 0, 0, 0]
[ 0, 0, 0, 0, 0, 0]
[ 0, 0, 0, 0, 0, 0]
"""
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matrix4()
Recall that if $f$ is an analytic function, then we can define $f(M)$ for any square matrix $M$ by "plugging" $M$ into the power series formula for $f(x)$. In other words, if $$f(x) = \sum_{n=0}^\infty a_n x^n,$$ then we define $f(M)$ by $$f(M) = \sum_{n=0}^\infty a_n M^n,$$ where $M^0$ is $I$, the identity matrix.
Furthermore, if $M$ is a diagonalizable matrix, that is, $M=PDP^{-1}$, where $D$ is diagonal, then $M^n = PD^nP^{-1}$ (because $M^n = \left(PDP^{-1}\right)\left(PDP^{-1}\right)\cdots\left(PDP^{-1}\right)=PD\left(P^{-1}P\right)D\left(P^{-1}P\right)\cdots DP^{-1} = PD^nP^{-1}$).
But if
$$ D = \begin{bmatrix} d_1 & 0 & \cdots & 0 \\\\ 0 & d_2 & \cdots & 0 \\\\ \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & \cdots & d_n \end{bmatrix} $$is a diagonal matrix, then
$$ D^n = \begin{bmatrix} d_1^n & 0 & \cdots & 0 \\\\ 0 & d_2^n & \cdots & 0 \\\\ \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & \cdots & d_n^n \end{bmatrix} $$so that
$$ \sum_{n=0}^\infty a_n M^n = \sum_{n=0}^\infty a_n PD^nP^{-1} = P\cdot\begin{bmatrix} \sum_{n=0}^\infty a_n d_1^n & 0 & \cdots & 0 \\\\ 0 & \sum_{n=0}^\infty a_n d_2^n & \cdots & 0 \\\\ \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & \cdots & \sum_{n=0}^\infty a_n d_n^n \end{bmatrix}\cdot P^{-1} = P\cdot\begin{bmatrix} f(d_1) & 0 & \cdots & 0 \\\\ 0 & f(d_2) & \cdots & 0 \\\\ \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & \cdots & f(d_n) \end{bmatrix}\cdot P^{-1} $$Let's create some square matrices, which we will use throughout the exercises.
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x = symbols('x')
A = Matrix([[1, 1], [1, 0]])
M = Matrix([[3, 10, -30], [0, 3, 0], [0, 2, -3]])
N = Matrix([[-1, -2, 0, 2], [-1, -1, 2, 1], [0, 0, 2, 0], [-1, -2, 2, 2]])
First, verify that these matrices are indeed diagonalizable.
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print(A.is_diagonalizable())
print(M.is_diagonalizable())
print(N.is_diagonalizable())
Now, we want to write a function that computes $f(M)$, for diagonalizable matrix $M$ and analytic function $f$.
However, there is one complication. We can use diagonalize to get P and D, but we need to apply the function to the diagonal of D. We might think that we could use eigenvals to get the eigenvalues of the matrix, since the diagonal values of D are just the eigenvalues of M, but the issue is that they could be in any order in D.
Instead, we can use matrix slicing to get the diagonal values (or indeed, any value) of a matrix. There is not enough time in this tutorial (or room in this document) to discuss the full details of matrix slicing. For now, we just note that M[i, j] returns the element at position i, j (which is the i + 1, j + 1th element of the matrix, due to Python's 0-indexing). For example
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M
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M[0, 1]
That should be enough information to write the following function.
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def matrix_func(M, func):
"""
Computes M at func. Assumes that M is square diagonalizable.
>>> matrix_func(M, exp)
[exp(3), -5*exp(-3)/3 + 5*exp(3)/3, -5*exp(3) + 5*exp(-3)]
[ 0, exp(3), 0]
[ 0, -exp(-3)/3 + exp(3)/3, exp(-3)]
Note that for the function exp, we can also just use M.exp()
>>> matrix_func(M, exp) == M.exp()
True
But for other functions, we have to do it this way.
>>> M.sin()
Traceback (most recent call last):
...
AttributeError: Matrix has no attribute sin.
>>> matrix_func(N, sin)
[-sin(1), -2*sin(1), 0, 2*sin(1)]
[-sin(1), -sin(1), sin(2), sin(1)]
[ 0, 0, sin(2), 0]
[-sin(1), -2*sin(1), sin(2), 2*sin(1)]
Note that we could also use this to compute the series expansion of a matrix,
if we know the closed form of that expansion. For example, suppose we wanted to compute
I + M + M**2 + M**3 + …
The series
1 + x + x**2 + x**3 + …
is equal to the function 1/(1 - x).
>>> matrix_func(M, Lambda(x, 1/(1 - x))) # Note, Lambda works just like lambda, but is symbolic
[-1/2, -5/4, 15/4]
[ 0, -1/2, 0]
[ 0, -1/4, 1/4]
"""
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matrix_func(M, exp)
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matrix_func(M, Lambda(x, 1/(1 - x)))
Now lets investigate how this works in relation to the series expansion definition. Write a function that uses matrix_func and series to compute the approximation of a matrix evaluated at a function up to $O(M^n)$.
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def matrix_func_series(M, func, n):
"""
Computes the approximation of the func(M) using the series definition up to O(M**n).
>>> matrix_func_series(M, exp, 10)
[22471/1120, 14953/448, -44859/448]
[ 0, 22471/1120, 0]
[ 0, 14953/2240, 83/2240]
>>> matrix_func_series(M, exp, 10).evalf()
[20.0633928571429, 33.3772321428571, -100.131696428571]
[ 0, 20.0633928571429, 0]
[ 0, 6.67544642857143, 0.0370535714285714]
>>> matrix_func(M, exp).evalf()
[20.0855369231877, 33.3929164246997, -100.178749274099]
[ 0, 20.0855369231877, 0]
[ 0, 6.67858328493993, 0.0497870683678639]
It's pretty close. Basically what we might expect for those values up to O(x**10).
>>> matrix_func_series(N, sin, 3)
[-1, -2, 0, 2]
[-1, -1, 2, 1]
[ 0, 0, 2, 0]
[-1, -2, 2, 2]
>>> matrix_func(N, sin).evalf()
[-0.841470984807897, -1.68294196961579, 0, 1.68294196961579]
[-0.841470984807897, -0.841470984807897, 0.909297426825682, 0.841470984807897]
[ 0, 0, 0.909297426825682, 0]
[-0.841470984807897, -1.68294196961579, 0.909297426825682, 1.68294196961579]
It's not as close, because we used O(x**3), but clearly still the same thing.
>>> matrix_func_series(M, Lambda(x, 1/(1 - x)), 10)
[29524, 73810, -221430]
[ 0, 29524, 0]
[ 0, 14762, -14762]
>>> matrix_func(M, Lambda(x, 1/(1 - x)))
[-1/2, -5/4, 15/4]
[ 0, -1/2, 0]
[ 0, -1/4, 1/4]
Woah! That one's not close at all. What is happening here? Let's try more terms
>>> matrix_func_series(M, Lambda(x, 1/(1 - x)), 100)
[257688760366005665518230564882810636351053761000, 644221900915014163795576412207026590877634402500, -1932665702745042491386729236621079772632903207500]
[ 0, 257688760366005665518230564882810636351053761000, 0]
[ 0, 128844380183002832759115282441405318175526880500, -128844380183002832759115282441405318175526880500]
It just keeps getting bigger. In fact, the series diverges. Recall that
1/(1 - x) = 1 + x + x**2 + x**3 + … *only if* |x| < 1. But the eigenvalues
of M are bigger than 1 in absolute value.
>>> M.eigenvals()
{3: 2, -3: 1}
In fact, 1/(1 - M) is mathematically defined via the analytic continuation
of the series expansion 1 + x + x**2 + …, which is just 1/(1 - x). This is
well-defined as long as none of the eigenvalues of M are equal to 1. Let's
try it on N.
>>> matrix_func(N, Lambda(x, 1/(1 - x)))
[nan, -oo, nan, oo]
[nan, nan, nan, nan]
[nan, nan, nan, nan]
[nan, -oo, nan, oo]
That didn't work. What are the eigenvalues of N?
>>> N.eigenvals()
{1: 1, 2: 1, -1: 1, 0: 1}
Ah, the first one is 1, so we cannot define 1/(1 - N).
"""
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matrix_func_series(M, exp, 10)
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matrix_func_series(M, exp, 10).evalf()
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matrix_func(M, exp).evalf()
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matrix_func_series(N, sin, 3)
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matrix_func_series(M, Lambda(x, 1/(1 - x)), 100)
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M.eigenvals()
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matrix_func(N, Lambda(x, 1/(1 - x)))
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N.eigenvals()
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