Confused on this defective matrix problem

[Dustinsfl]

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

\begin{bmatrix}<br /> 3\alpha &amp; 1 &amp; 0\\ <br /> 0 &amp; \alpha &amp; 0\\ <br /> 0 &amp; 0 &amp; \alpha<br /> \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}<br /> 2\alpha &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 1 &amp; \frac{-1}{2\alpha} &amp; 0\\ <br /> 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0<br /> \end{bmatrix}\Rightarrow x_2\begin{bmatrix}<br /> \frac{-1}{2\alpha}\\ <br /> 1\\ <br /> 0 <br /> \end{bmatrix}+x_3\begin{bmatrix}<br /> 0\\ <br /> 0\\ <br /> 1<br /> \end{bmatrix}

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

For lambda being 3alpha
\begin{bmatrix}<br /> 0 &amp; 1 &amp; 0\\ <br /> 0 &amp; -2\alpha &amp; 0\\ <br /> 0 &amp; 0 &amp; -2\alpha<br /> \end{bmatrix}\Rightarrow \begin{bmatrix}<br /> 0 &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 1\\ <br /> 0 &amp; 0 &amp; 0<br /> \end{bmatrix}\Rightarrow x_1\begin{bmatrix}<br /> 1\\ <br /> 0\\ <br /> 0<br /> \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?