##### Example4.2.6

We saw earlier that the matrix $$\left[\begin{array}{rrr} 3 \amp -1 \amp 4 \\ 0 \amp -2 \amp 3 \\ 0 \amp 0 \amp 1 \\ \end{array}\right]$$ has the characteristic equation

\begin{equation*} (3-\lambda)(-2-\lambda)(1-\lambda)=0\text{.} \end{equation*}

There are three eigenvalues $$\lambda=3,-2,1$$ each having multiplicity $$1\text{.}$$ By the proposition, we are guaranteed that the dimension of each eigenspace is $$1\text{;}$$ that is,

\begin{equation*} \dim E_3 = \dim E_{-2} = \dim E_1 = 1 \text{.} \end{equation*}

It turns out that this is enough to guarantee that there is a basis of $$\real^3$$ consisting of eigenvectors.

in-context