1

Determine the number of possible values for $\det(A)$, given that $A$ is an $n \times n$ matrix with real entries such that $A^3 - A^2 -3A +2I=0$.

here is the source of the problem. In the last comment, I don't understand why we can say for sure that $A$ has only three distinct eigenvalues. Can anyone explain to me?

Remark: I understand that the roots of the equation $x^3-x^2-3x+2=0$ are eigenvalues of $A$. My problem is why the matrix $A$ cannot have another eigenvalue?

Martin Sleziak
  • 56,060
Idonknow
  • 16,493
  • For your last remark, suppose $\lambda$ is an eigenvalue and $v\neq 0$ is an associated eigenvector, then $0=(A^3-A^2-3A+2I)v=(\lambda^3-\lambda^2-3\lambda+2)v$ which implies $\lambda^3-\lambda^2-3\lambda+2=0$. To be even more explicit, if $\lambda$ is eigenvalue, then $\lambda$ necessarily is a root of $z^3-z^2-3z+2$. – Kim Jong Un May 20 '15 at 12:11
  • Note that the minimal polynomial of $A$ has the exact same roots as its characteristic polynomial – gebruiker May 20 '15 at 12:12
  • 1
    http://math.stackexchange.com/questions/101275/roots-of-minimal-and-characteristic-polynomial – Martin Sleziak May 20 '15 at 12:35

3 Answers3

3

Hint : The minimal polynomial of the matrix $A$ must be a divisor of the polynomial

$$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$

SLeslie
  • 2,742
Peter
  • 86,576
  • Indeed, if $(x,v)$ is an eigenpair of $A$ there holds $$ (A^3-A^2-3A+2I)v=(x^3-x^2-3x+2)x~v=(x-2)(x^2+x-1)~v. $$

    Since the roots of $(x-2)(x^2+x-1)$ are real and distinct, there holds that $A$ is diagonalizable and whence, the determinant of $A$ is of the form $2^i (\phi_+)^j (\phi_-)^k$, with $i+j+k=n$ ($i,j,k\geq 1$).

    Here and elsewhere $\phi_\pm=\dfrac{1\pm \sqrt{5}}{2}$.

     – Nelson Faustino Apr 20 '19 at 18:46
2

Suppose $v$ is an eigenvector of $A$, with eigenvalue $\lambda$. Then $\lambda$ must satisfy the same cubic equation that $A$ satisfies: $$(A^3-A^2-3A+2I)v=0v=0\\ (A^3-A^2-3A+2I)v=AAAv-AAv-3Av+2v\\ =AA\lambda v-A\lambda v-3\lambda v+2v\\ =\lambda A(Av)-\lambda(Av)-3\lambda v+2v\\ =\lambda A(\lambda v)-\lambda^2 v-3\lambda v+2v\\ =\lambda^2(Av)-\lambda^2v-3\lambda v+2v =(\lambda^3-\lambda^2-3\lambda+2)v$$ From the first line, the answer is zero, but $v$ is not the zero vector, so $\lambda^3-\lambda^2-3\lambda+2=0$

Empy2
  • 52,372
1

The eigenvalues of $A$ are the roots of its minimal polynomial, and $\bigl\{P(x)\in \mathbf R[x] \mid P(A)=0\bigr\}$ is the set of multiples of the minimal polynomial of $A$. Hence if $P(A)=0$, the eigenvalues of $A$ are among the roots of $P(x)$.

Bernard
  • 179,256