Mathematica Asked on October 1, 2021
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).
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]
Correct answer by Nasser on October 1, 2021
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]
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}]
See also comment @Szabolcs: $I^2+ I =2 I$ for any $n$ as well as $(-2I)^2+(-2 I)=2I$
Answered by ubpdqn on October 1, 2021
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