- #1

- 699

- 5

1 & 1 & 0\\

1 & 1 & 0\\

0 & 0 & \alpha

\end{bmatrix}[/tex] and [itex]det(A-\lambda I)=\lambda(\alpha-\lambda)(\lambda-2)=0[/itex]

Therefore, [itex]\lambda_1=0[/itex], [itex]\lambda_2=\alpha[/itex], and [itex]\lambda_3=2[/itex].

In order to make this matrix defective, I need to make the left two column vectors dependent[tex]\begin{bmatrix}

1-\alpha & 1 & 0\\

1 & 1-\alpha & 0\\

0 & 0 & 0

\end{bmatrix}[/tex].

Is there an easy way to find or determine if this can be done for a given matrix?