Confused on this defective matrix problem

[Dustinsfl]

Make matrix defective if possible and identify the values of alpha.

$$\begin{bmatrix} 3\alpha & 1 & 0\\ 0 & \alpha & 0\\ 0 & 0 & \alpha \end{bmatrix}$$

Skipping the boring stuff we obtain $(\alpha-\lambda)^2(3\alpha-\lambda)=0$ as the characteristic polynomial.

$$\lambda_1=\lambda_2=\alpha$$ and $$\lambda_3=3\alpha$$
For lambda being alpha
$$\begin{bmatrix} 2\alpha & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & \frac{-1}{2\alpha} & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\Rightarrow x_2\begin{bmatrix} \frac{-1}{2\alpha}\\ 1\\ 0 \end{bmatrix}+x_3\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$

For this lambda value, the matrix can't be defective? Not sure though.

For lambda being 3alpha
$$\begin{bmatrix} 0 & 1 & 0\\ 0 & -2\alpha & 0\\ 0 & 0 & -2\alpha \end{bmatrix}\Rightarrow \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}\Rightarrow x_1\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$$

And for this one, alpha can be any value and the matrix will be defective.

 

[gabbagabbahey]

A 3x3 matrix is defective it has fewer that 3 distinct, linearly independent eigenvectors. How many distinct, linearly independent eigenvectors does this matrix have?