# Thread: Make a Matrix Defective

1. ## Make a Matrix Defective

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?

2. 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?
A matrix is defective if there are any repeated eigenvalues.

So you have

$\displaystyle \left|\begin{matrix} 1-\lambda & 1 & 0\\ 1 & 1-\lambda & 0\\ 0 & 0 & \alpha-\lambda \end{matrix}\right|$

$\displaystyle = (\alpha - \lambda)\left|\begin{matrix} 1-\lambda & 1\\ 1 & 1-\lambda\end{matrix}\right|$

$\displaystyle = (\alpha - \lambda)[(1 - \lambda)^2 - 1^2]$

$\displaystyle = (\alpha - \lambda)(1 - \lambda + 1)(1 - \lambda - 1)$

$\displaystyle = -\lambda(\alpha - \lambda)(2 - \lambda)$.

So to have repeated eigenvalues, then $\displaystyle \alpha = 2$ or $\displaystyle \alpha = 0$.

3. Solved