Originally Posted by

**dwsmith** Find all possible values for $\displaystyle \alpha$ that make the matrix defective or show that no such value exist.

$\displaystyle \begin{bmatrix}

1 & 1 & 0\\

1 & 1 & 0\\

0 & 0 & \alpha

\end{bmatrix}\Rightarrow \begin{vmatrix}

1-\lambda & 1 & 0\\

1 & 1-\lambda & 0\\

0 & 0 & \alpha-\lambda

\end{vmatrix}=\lambda(\alpha-\lambda)(\lambda-2)=0$

Therefore, $\displaystyle \lambda_1=0$, $\displaystyle \lambda_2=\alpha$, and $\displaystyle \lambda_3=2$.

$\displaystyle \begin{bmatrix}

1-\alpha & 1 & 0\\

1 & 1-\alpha & 0\\

0 & 0 & 0

\end{bmatrix}$ what I am thinking is that this can only occur if $\displaystyle \alpha=0$ or $\displaystyle 2$. Then if that doesn't work, there is no such value; however, I am not sure if that thought is entirely true. Would it be possible for other values of alpha to make the matrix defective?