Question

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.

Answer 1

With

`\mathbf A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}`

we have

`\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`

so
`\mathbf A` has one eigenvalue,
`\lambda=2`, with multiplicity 3.

In order for
`\mathbf A` to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But
`\mathbf A-2\mathbf I` has nullspace of dimension 2, since

`\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`

That is, we can only obtain 2 eigenvectors,

`\begin{bmatrix}1\\0\\0\end{bmatrix}\text{ and }\begin{bmatrix}0\\0\\1\end{bmatrix}`

and there is no other. We needed 3 in order to complete the basis of eigenvectors.