# Simplifying away imaginary part

I'm solving the following eigensystem, and I get result which looks complex-valued. I expect the result to have 0 imaginary part, can anyone see a way to simplify it away?

B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/
3}, {1/3, 1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3,
0, 10/3}};
First /@ Eigensystem[{B, A}, 1]


B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3,
1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3, 0,
10/3}};

val = First /@ Eigensystem[{B, A}, 1];


All of the values are real

valR = val // RootReduce


Element[valR, Reals]

(* True *)

valR // N

(* {8.33542, {0.0311157, -0.179391, 0.00719285, 1.}} *)


However, if represented using radicals, they must use complex numbers (see Casus irreducibilis)

valR // ToRadicals


• aha, RootReduce was the magic keyword! Commented Sep 11, 2020 at 3:36
Chop[N[First /@ Eigensystem[{B, A}, 1]]]


returns

{8.33542,{0.0311157,-0.179391,0.00719285,1.}}

• I'm not 100% sure the imaginary part is exactly 0, or just close to 0, so was trying to use Mathematica to confirm my expectation Commented Sep 11, 2020 at 0:01
• N[First /@ Eigensystem[{B, A}, 1],256] tries to give you the first 256 digits of each number and that shows if there is an imaginary part that it is preceded by about 256 zeros.
– Bill
Commented Sep 11, 2020 at 1:55

This takes a long while, but it works if you just want to prove that the imaginary part is $$0$$.

B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/
3}, {1/3, 1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3,
0, 10/3}};
result = First /@ Eigensystem[{B, A}, 1]

Im[result] // FullSimplify
(*{0,{0,0,0,0}}*)


I let the computer run overnight to get the answer. Simplify did not get there.

• I'm eagerly waiting for computers to get fast enough for FullSimplify to run faster than overnight Commented Sep 11, 2020 at 18:49
• FullSimplify@Im@Numerator@Together@ComplexExpand@result will wrap up before you can fall asleep ;) Commented Sep 11, 2020 at 20:08

Try numerical evaluation with Mathematica's N[]

B = {{17/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3,
1/3}, {1/3, 1/3, 1/3, 82/3}};
A = {{5/3, 0, 1/3, 0}, {0, 5/3, 0, 1/3}, {1/3, 0, 10/3, 0}, {0, 1/3,
0, 10/3}};
First /@ Eigensystem[{N@B, N@A}, 1]


The result should be

{8.33542, {0.0306117, -0.176485, 0.00707634, 0.983802}}

• interesting, it seems applying N before Eigenvalues is better than applying N after Commented Sep 11, 2020 at 4:09

To piggyback off Bill's answer, one can just use CountRoots[] on the characteristic polynomial of the given matrix pencil, if one only wishes to show that the eigenvalues are all real:

CountRoots[CharacteristicPolynomial[{B, A}, x], x]
4


One can then use RootIntervals[] to find brackets for the roots:

RootIntervals[CharacteristicPolynomial[{B, A}, x], Reals]
{{{0, 0}, {0, 1}, {3, 4}, {4, 10}}, {{1}, {1}, {1}, {1}}}


Note that the root at $$x=0$$ was exactly isolated. The largest eigenvalue of the pencil would correspond to the last entry with the isolating interval $$(4,10)$$, which you can then give to Solve[]:

Solve[CharacteristicPolynomial[{B, A}, x] == 0 && 4 < x < 10, x,
Cubics -> False, Quartics -> False]
{{x -> Root[-19440 + 76898 #1 - 28959 #1^2 + 2401 #1^3 &, 3]}}


Bob has already mentioned casus irreducibilis; to summarize, if you insist on a radical representation, then the use of a complex representation is (often) unavoidable, even if all the roots are real.