To be diagonalizable means to be conjugate to a diagonal matrix. That is, A is diagonalizable if there is a diagonal matrix D and an invertible matrix P such that
P^(-1) A P = D
Another way to say this is that A is similar to D. Many important properties are always shared by similar matrices. For example, rank. The rank of any two similar matrices is equal. Another one is their determinant.
Another one is their characteristic polynomial. Since they share the same polynomial, they necessarily share the same eigenvalues. Finally, suppose A and D are similar, and they each have the eigenvalue lambda. Then each has an associated eigenspace associated to lambda, and we can say that the dimension of these eigenspaces are equal (we cannot say that the eigenspaces are equal, only the dimension).
Now, stare at any diagonal 3 x 3 matrix D for a second, and call its diagonal entries d1,d2, and d3. You should notice that each of its diagonal entries are eigenvalues for D, and that an eigenvector for d1 is [1,0,0], for d2 is [0,1,0], for d3 is [0,0,1]. Thus D has three linearly independent eigenvectors.
So, if A is diagonalizable, A must also have three linearly independent eigenvectors. As it turns out, this is an if and only if statement. That is,
The n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.
(One can prove this directly... let your transition matrix P be the matrix whose columns consist of A's n linearly independent eigenvectors. Then P^(-1) A P will be a diagonal matrix D whose diagonal entries are the eigenvectors of A).
So: does your matrix A have three linearly independent eigenvectors?