The main issue is that you deal with symbolic expressions (where computations are hard, for example checking that some symbolic expression is equal to zero is undecidable), not algebraic or numerical ones:

```
sage: k=6 ; m=a=b=c=1
sage: w = exp((2*pi*I*m )/k)
sage: A = matrix([[0,1,w^a,1],[1,0,1,w^(k-b)],[w^(k-a),1,0,w^(k-c)],[1,w^b,w^c,0]])
sage: A
[ 0 1 e^(1/3*I*pi) 1]
[ 1 0 1 e^(5/3*I*pi)]
[e^(5/3*I*pi) 1 0 e^(5/3*I*pi)]
[ 1 e^(1/3*I*pi) e^(1/3*I*pi) 0]
sage: A.parent()
Full MatrixSpace of 4 by 4 dense matrices over Symbolic Ring
```

You can compute the characteristic polynomial:

```
sage: P = A.characteristic_polynomial()
sage: P
x^4 - 6*x^2 - 6*x - 1
```

But it has symbolic coefficients, so it is not able to find its roots:

```
sage: P.parent()
Univariate Polynomial Ring in x over Symbolic Ring
sage: P.roots()
NotImplementedError:
```

What you should do is to specify a better ring for the coefficients. You will see that the roots depend on the ring:

There is one integer root with multiplicity 1:

```
sage: Q = P.change_ring(ZZ)
sage: Q.roots()
[(-1, 1)]
```

There is one rational root with multiplicity 1:

```
sage: Q = P.change_ring(QQ)
sage: Q.roots()
[(-1, 1)]
```

They are 4 algebraic roots, each with multipicity 1.

```
sage: Q = P.change_ring(QQbar)
sage: Q.roots()
[(-1.655442381549831?, 1),
(-1, 1),
(-0.2107558809591918?, 1),
(2.866198262509023?, 1)]
```

You can also work in inexact rings, to get floating point approximations, for example in the `Real Double Field`

:

```
sage: Q = P.change_ring(RDF)
sage: Q.roots()
[(-1.6554423815498323, 1),
(-0.9999999999999978, 1),
(-0.21075588095919173, 1),
(2.866198262509024, 1)]
```

Note that there is a huge difference between the last two blocks, the one with `QQbar`

gives you exact results (if you evaluate the polynomial with them you will get a genuine zero), the one with `RDF`

gives you approximate result.

**EDIT** : In your particular case, it seems that Sage is able to find the roots of the symbolic polynomial:

```
sage: solve(P(x), x)
[x == -1/2*(1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3)*(I*sqrt(3) + 1) + 1/9*(8*I*sqrt(3) - 8)/(1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3) + 1/3, x == -1/2*(1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3)*(-I*sqrt(3) + 1) + 1/9*(-8*I*sqrt(3) - 8)/(1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3) + 1/3, x == (1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3) + 16/9/(1/9*I*sqrt(101)*sqrt(3) + 37/27)^(1/3) + 1/3, x == -1]
```

However, the result is far from being trustable and i would not recommend to use it in general, for example there will be situations where `solve ...`

(more)