evalf() returns symbolic expression or sum of complex conjugates

[step135]

M=sqrt(Matrix([
 [ 1, 1],
 [-1, 1]]))
print (M.evalf())

sqrt is not evaluated symbolically and neither evalf() is transforming it to numbers.

Compared to:

M=Matrix([
 [ 1, 1],
 [-1, 1]])**(1./2)
print (M.evalf())

Output here contains (1 - I)**0.5/2 + (1 + I)**0.5/2. That should be automatically simplified as it is possible: = sqrt(sqrt(2))*cos(pi/8) = sqrt(sqrt(2))*sqrt(sqrt(2)/4 + 1/2)

So according to examples there are three not related bugs by themselves:

1. sqrt on Matrix doesn't count symbolic result, but **(1./2) does it
2. evalf() doesn't transform symbolic expressions into numbers in some cases
3. evalf() is providing sum of complex conjugates that is easy to simplify in many cases into real number, but it shows more complicated complex result

[oscarbenjamin]

You need doit for the symbolic matrix square root:

In [29]: M=sqrt(Matrix([ 
    ...:  [ 1, 1], 
    ...:  [-1, 1]]))                                                                                                                           

In [30]: M                                                                                                                                     
Out[30]: 
         1/2
⎛⎡1   1⎤⎞   
⎜⎢     ⎥⎟   
⎝⎣-1  1⎦⎠   

In [31]: M.doit()                                                                                                                              
Out[31]: 
⎡     _______     _______           _______       _______⎤
⎢   ╲╱ 1 - ⅈ    ╲╱ 1 + ⅈ        ⅈ⋅╲╱ 1 + ⅈ    ⅈ⋅╲╱ 1 - ⅈ ⎥
⎢   ───────── + ─────────     - ─────────── + ───────────⎥
⎢       2           2                2             2     ⎥
⎢                                                        ⎥
⎢      _______       _______       _______     _______   ⎥
⎢  ⅈ⋅╲╱ 1 - ⅈ    ⅈ⋅╲╱ 1 + ⅈ      ╲╱ 1 - ⅈ    ╲╱ 1 + ⅈ    ⎥
⎢- ─────────── + ───────────     ───────── + ─────────   ⎥
⎣       2             2              2           2       ⎦

In [32]: M.doit().rewrite(exp)                                                                                                                 
Out[32]: 
⎡          -ⅈ⋅π           ⅈ⋅π                ⅈ⋅π            -ⅈ⋅π ⎤
⎢          ─────          ───                ───            ─────⎥
⎢   4 ___    8     4 ___   8        4 ___     8    4 ___      8  ⎥
⎢   ╲╱ 2 ⋅ℯ        ╲╱ 2 ⋅ℯ          ╲╱ 2 ⋅ⅈ⋅ℯ      ╲╱ 2 ⋅ⅈ⋅ℯ     ⎥
⎢   ──────────── + ──────────     - ──────────── + ──────────────⎥
⎢        2             2                 2               2       ⎥
⎢                                                                ⎥
⎢           -ⅈ⋅π             ⅈ⋅π            -ⅈ⋅π           ⅈ⋅π   ⎥
⎢           ─────            ───            ─────          ───   ⎥
⎢  4 ___      8     4 ___     8      4 ___    8     4 ___   8    ⎥
⎢  ╲╱ 2 ⋅ⅈ⋅ℯ        ╲╱ 2 ⋅ⅈ⋅ℯ        ╲╱ 2 ⋅ℯ        ╲╱ 2 ⋅ℯ      ⎥
⎢- ────────────── + ────────────     ──────────── + ──────────   ⎥
⎣        2               2                2             2        ⎦

Unfortunately it isn't possible to rewrite that as cos because cos(pi/8) evaluates automatically:

In [36]: exp(I*pi/8)                                                                                                                           
Out[36]: 
 ⅈ⋅π
 ───
  8 
ℯ   

In [37]: exp(I*pi/8).rewrite(cos)                                                                                                              
Out[37]: 
    ________         ________
   ╱ √2   1         ╱ 1   √2 
  ╱  ── + ─  + ⅈ⋅  ╱  ─ - ── 
╲╱   4    2      ╲╱   2   4  

In [38]: cos(pi/8)                                                                                                                             
Out[38]: 
    ________
   ╱ √2   1 
  ╱  ── + ─ 
╲╱   4    2 

[step135]

Thanks, I just realized that sqrt compared to ** doesn't evaluate symbolically.

So points 2) and 3) refers still to bugs.
To 2) evalf() should evaluate expression even without doit().
To 3) the matrix is diagonalizable so there is sqrt on eigenvalues that are complex.

Sympy doesn't symbolically evaluate expressions of types (1+i)**(1./2). Eigenvalues are 1 + i and 1 - i.

It could learn to solve it automatically as = (sqrt(2)*exp(i*atan(1))**(1./2) = (sqrt(2)*exp(i*pi/4))**(1./2) = sqrt(sqrt(2))*exp(i*pi/8), because it knows all following identities needed: atan(1) = pi/4, exp(i*pi/4) = sqrt(2)/2*(1+i) and also cos(2*x) = 2*cos(x)^2 - 1