How to find the eigenvalues and eigenvectors of a matrix with SymPy?

Question

I want to calculate the eigenvectors x from a system A by using this: A x = λ x

The problem is that I don't know how to solve the eigenvalues by using SymPy. Here is my code. I want to get some values for x1 and x2 from matrix A

from sympy import *
x1, x2, Lambda = symbols('x1 x2 Lambda')
I = eye(2)
A = Matrix([[0, 2], [1, -3]])
equation = Eq(det(Lambda*I-A), 0)
D = solve(equation)
print([N(element, 4) for element in D]) # Eigenvalus in decimal form
print(pretty(D)) # Eigenvalues in exact form

X = Matrix([[x1], [x2]]) # Eigenvectors
T = A*X - D[0]*X # The Ax = %Lambda X with the first %Lambda = D[0]
print(pretty(solve(T, x1, x2)))

score 9 · Answer 1 · answered Apr 28 '17 at 22:52

The methods eigenvals and eigenvects is what one would normally use here.

A.eigenvals() returns {-sqrt(17)/2 - 3/2: 1, -3/2 + sqrt(17)/2: 1} which is a dictionary of eigenvalues and their multiplicities. If you don't care about multiplicities, use list(A.eigenvals().keys()) to get a plain list of eigenvalues.

The output of eigenvects is a bit more complicated, and consists of triples (eigenvalue, multiplicity of this eigenvalue, basis of the eigenspace). Note that the multiplicity is algebraic multiplicity, while the number of eigenvectors returned is the geometric multiplicity, which may be smaller. The eigenvectors are returned as 1-column matrices for some reason...

For your matrix, A.eigenvects() returns the eigenvector [-2/(-sqrt(17)/2 + 3/2), 1] for the eigenvalue -3/2 + sqrt(17)/2, and eigenvector [-2/(3/2 + sqrt(17)/2), 1] for eigenvalue -sqrt(17)/2 - 3/2.

If you want the eigenvectors presented as plain lists of coordinates, the following

[list(tup[2][0]) for tup in A.eigenvects()]

would output [[-2/(-sqrt(17)/2 + 3/2), 1], [-2/(3/2 + sqrt(17)/2), 1]]. (Note this just picks one eigenvector for each eigenvalue, which is not always what you want)

Hi! Yes, i Know how to get the eigenvects by using the built in functions. But if I want to use the solve-function. As you see, it's not working for me. I cannot calculate out x1 and x2 which is the eigenvectors. — euraad, Apr 29 '17 at 07:09
The above line doesnt always work, for example for this matrix there are 3 Eign vectors and that line above shows me only 2. ``` A = Matrix([ [4, 0, 1], [2, 3, 2], [1, 0, 4] ]) ``` — Mohammed, Nov 14 '20 at 17:52
This line worked for me best ``` sym_eignvects = [] for tup in sMatrix.eigenvects(): for v in tup[2]: sym_eignvects.append(list(v)) ``` — Mohammed, Nov 14 '20 at 18:19

score 7 · Accepted Answer · answered May 26 '17 at 14:37

7

sympy has a very convenient way of getting eigenvalues and eigenvectors: sympy-doc

Your example would simply become:

from sympy import *
A = Matrix([[0, 2], [1, -3]])
print(A.eigenvals())  #returns eigenvalues and their algebraic multiplicity
print(A.eigenvects())  #returns eigenvalues, eigenvects

answered May 26 '17 at 14:37

laolux

1,445
1
17
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2

[T, D] = A.diagonalize() was a better method in SymPy. – euraad May 27 '17 at 19:27

score 0 · Answer 3 · answered Nov 14 '20 at 18:21

This answer will help you when you all eignvectors, the solution above doesnt always give you all eienvectos for example this matrix A used below

# the matrix
A = Matrix([
    [4, 0, 1],
    [2, 3, 2],
    [1, 0, 4]
])


    sym_eignvects = []
    for tup in sMatrix.eigenvects():
        for v in tup[2]:
            sym_eignvects.append(list(v))

How to find the eigenvalues and eigenvectors of a matrix with SymPy?

3 Answers3