Prove that A is diagonalizable

Question

Given matrix $A_{3x3}$ , and $Ax = 0$, $Ax = x$, $Ax = -x$ all have non trivial solutions. How can I prove that $A$ is diagonalizable?

As a start I know that $A$ is not invertible (concluded from $Ax = 0$ has a non trivial solution), the determinant of $A$ is zero. But what does the other information give me?

Answer 1

You have three distinct eigenvalues in a $3\times 3$ matrix. That means that you have three linearly independent eigenvectors in $\Bbb R^3$. In that basis $A$ is diagonal.

Answer 2

I'm assuming A is defined over a field F with characteristic different from 2.

The conditions imply that A has eigenvalues 0, 1 and -1, respectively. Different eigenvalues implies that the corresponding eigenvectors are linearly independent. Since A is 3x3, this means that these eigenvectors form a basis of the vector space F^3. Since there is a basis of eigenvectors, A is diagonalizable.

In particular, A is similar to diag(0,1,-1).

Answer 3

If eigenvectors $v_1,\dots, v_r$ correspond to distinct eigenvalues $\alpha_1,\dots, \alpha_r$, the set $\bigl\{v_1,\dots, v_r\bigr\}$ is linearly independent.

Hence, if you have $n$ distinct eigenvalues for an $n\times n$ matrix $A$, representing an endomorphism of the space $K^n$ ($K$ denotes the base field), this matrix is diagonalisable to a matrix $D=\operatorname{Diag}(\alpha_1, \dots,\alpha_n) $ in a basis of $n$ eigenvectors.