Linear Algebra-Eigenvalues

Question

Suppose an $n*n$ matrix $A$ has the property that $A^{3}=A$. Show that the only possible eigenvalues of $A$ are $-1,0$ and $1$.

Answer 1

If $\lambda$ is an eigenvalue of $A$ and $\vec v \ne 0$ is an associated eigenvector, then

$A \vec v = \lambda \vec v, \tag 1$

whence

$A^3 \vec v = A^2(A \vec v) = A^2 (\lambda \vec v) = \lambda (A^2 \vec v) = \lambda A (A \vec v) = \lambda^2 A \vec v = \lambda^3 \vec v. \tag 2$

Since

$A^3 = A, \tag 3$

we have

$\lambda^3 \vec v = A^3 \vec v = A \vec v = \lambda \vec v; \tag 4$

thus

$(\lambda^3 - \lambda) \vec v = 0, \tag 5$

which since $\vec v \ne 0$ forces

$\lambda (\lambda + 1) (\lambda - 1) = \lambda^3 - \lambda = 0; \tag 6$

we see from the factorization given in (6) that the only possible values for $\lambda$ are

$\lambda = 0, -1, 1. \tag 7$

Answer 2

Hint: If $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector, what can you say about $A.v$? And about $A^3.v$?

Answer 3

Note that

$$Av=\lambda v\implies A^2v=\lambda Av=\lambda^2v\implies A^3v=Av=\lambda^3v=\lambda v$$

thus

$$\lambda^3=\lambda\implies\lambda^3-\lambda=0\implies\lambda(\lambda^2-1)=0\implies\lambda=0,1,-1$$

Answer 4

In the diagonalized form,

$$A=PDP^T$$ and $$A^3=PD^3P^T,$$

so that $$D^3=D.$$

Hence the Eigenvalues verify $\lambda^3=\lambda$.