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It is known that $E$ is a third-order identity matrix and $A$ is a third-order real symmetric matrix. Matrix $A$ satisfies the condition of $A^{2}+A=2 E$.

How to find the eigenvalue of this abstract matrix (reference answer: -2,-2,1).

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    $\begingroup$ This seems to be more a question about mathematics than about Mathematica. $\endgroup$ – Michael Seifert Jul 29 at 15:25
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    $\begingroup$ @MichaelSeifert On one hand yes, but it can be easily understood as "how to do it entirely in Mathematica as a demonstration", which is how it turned out in Nasser's answer. $\endgroup$ – corey979 Jul 29 at 17:48
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How to find the eigenvalue of this abstract matrix (reference answer: -2,-2,1).

I am assuming $A^2$ means $A*A$? If so, then may be

e = IdentityMatrix[3];
a = {{a11, a21, a31}, {a21, a22, a32}, {a31, a32, a33}}(*abstract symmetric matrix*)
eqs = Thread[Flatten[a.a + a] == Flatten[2*e]];
sol = FindInstance[eqs, Flatten[a]];
Eigenvalues[a /. sol]

Mathematica graphics

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This suggests minimal polynomial of $A$ is $x^2+x-2$. This has roots -2,1 (the eigenvalues not counting multiplicities). Note $m(A)$ divides $p(A)$ (the characteristic polynomial, which by Cayley-Hamilton $p(A)=0$)

(-2,-2,1) or (-2,1,1) could be eigenvalues of $A$: e.g. using diagonal matrices. This also case when examine other instances from @Nasser code.

For illustrative purposes:

Solve[x^2 + x - 2 == 0, x]
a1 = {{-2, 0, 0}, {0, -2, 0}, {0, 0, 1}};
a2 = {{-2, 0, 0}, {0, 1, 0}, {0, 0, 1}};
Eigenvalues[a1]
Eigenvalues[a2]
a1.a1 + a1 // MatrixForm
a2.a2 + a2 // MatrixForm
MatrixMinimalPolynomial[a_List?MatrixQ, x_] := 
 Module[{i, n = 1, qu = {}, 
   mnm = {Flatten[IdentityMatrix[Length[a]]]}}, 
  While[Length[qu] == 0, AppendTo[mnm, Flatten[MatrixPower[a, n]]];
   qu = NullSpace[Transpose[mnm]];
   n++];
  First[qu].Table[x^i, {i, 0, n - 1}]]
MatrixMinimalPolynomial[a1, x]
MatrixMinimalPolynomial[a2, x]

enter image description here

Illustrating from @Nasser code:

e = IdentityMatrix[3];
a = {{a11, a21, a31}, {a21, a22, a32}, {a31, a32, 
   a33}};(*abstract symmetric matrix*)
eqs = Thread[Flatten[a.a + a] == Flatten[2*e]];
sol = FindInstance[eqs, Flatten[a], 10];
Table[Eigenvalues[a /. sol[[j]]], {j, 10}]

enter image description here

See also comment @Szabolcs: $I^2+ I =2 I$ for any $n$ as well as $(-2I)^2+(-2 I)=2I$

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    $\begingroup$ $(1,1,1)$ is also possible if $A$ is the identity matrix. The result is a bit ugly, but Solve instead of FindInstance does succeed with Nasser's code. $\endgroup$ – Szabolcs Jul 29 at 8:38
  • $\begingroup$ @Szabolcs thank you. Of course you are correct. $\endgroup$ – ubpdqn Jul 29 at 8:40

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