Exercise 3.7.4: let a = 2 1 0 0 2 0 0 0 2 .a.what are the eigenvalues?b.what is/are the defect(s) of the eigenvalue(s)?c.find the general solution of ~x 0 = a~x in two di erent ways and verify you get the same answer.

With[tex]\mathbf A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}[/tex]we have[tex]\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&0&2-\lambda\end{vmatrix}=(2-\lambda)^3[/tex]so [tex]\mathbf A[/tex] has one eigenvalue, [tex]\lambda=2[/tex], with multiplicity 3.In order for [tex]\mathbf A[/tex] to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But [tex]\mathbf A-2\mathbf I[/tex] has nullspace of dimension 2, since[tex]\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\mathbf 0\implies x=0\text{ or }y=0[/tex]That is, we can only obtain 2 eigenvectors,[tex]\begin{bmatrix}1\\0\\0\end{bmatrix}\text{ and }\begin{bmatrix}0\\0\\1\end{bmatrix}[/tex]and there is no other. We needed 3 in order to complete the basis of eigenvectors.