# Jordan Canonical form $3\times3$

Let $$A= \begin{pmatrix} 3& 1& -2 \\ -1& 0& 5 \\ -1& -1& 4\end{pmatrix}$$ express it as Jordan Canonical form.

This is an example in book by Friedberg,Insel,Spence 2 on sec 7.1. Was able to follow that characteristic poly of A is $$f(t)=det(A-tI_{3x3})=-1(t-3)(t-2)^2$$ let $\lambda_1=3$ with mult 1 and $\lambda_2=2$ with mult 2. by thm $dim(K_{\lambda_1}(T))=1$ and $dim(K_{\lambda_2}(T))=2$

and by some proposition \begin{aligned} K_{\lambda_1}&=N(T-3I)=E_{\lambda_1} \\K_{\lambda_2}&=N((T-2I)^2) &&[\text{the book has maybe 1 ed. older }N((T-3I)^2) ] \end{aligned} I was able to find nullpace $N(T-3I)$ by rref it is $\{ (-1,2,1)\}=K_{\lambda_1}$.

By rref $N(T-2I)$ to be $\{ -1,3,-1\}$

cannot easily verify the following when it comes to finding $E_{\lambda_2}$

the generalized eigenspace has a basis consisting of a union of cycles, the basis is either union of 2 cycles each length 1 or a single cycle of length 2. The former case is impossible because the basis elements would be eigevectors- contradicting the fact that dim$(E_{\lambda_2})=1$ which is easily verified.

now it stated $$\beta_2=\{ (A-2I)v,v\} =\left \{ \begin{pmatrix} 1\\-3\\1 \end{pmatrix} , \begin{pmatrix}1\\2\\0 \end{pmatrix} \right \}$$ Not sure how $(1,2,0)$ was derived. besides doing inverse

• The characteristic polynomial of a 3x3 matrix should have degree 3, you've written down something that has degree 4. – Jim Apr 14 '16 at 21:39
• @jim had $-t$ it was $-1$. Fixed it. thanks – Tiger Blood Apr 14 '16 at 21:42

Let $B = A - 2I$. What you want is a vector $v$ so that $v$ and $Bv$ are nonzero but $B^2v = 0$. You know that the nullspace of $B^2$ is dimension $2$ and the nullspace of $B$ is dimension $1$, so pick $v$ to be any vector in $N(B^2) \setminus N(B)$. The vector $(1, 2, 0)$ is one of many possible choices.