Showing that the limit of non-eigenvector goes to infinity Let $A$ be a $3$ by $3$ real matrix with the triple eigenvalue $1$. Also, further suppose its eigenspace corresponding to $1$ is only of dimension $1$. Thus, we can find a basis of $\mathbb{R}^3$, denoted by $v$. $w_1$. $w_2$ where $v$ is an eigenvector of $A$.
Then I have to show that $\lim_{n \to \infty} \|A^n w_1\|=\lim_{n \to \infty} \|A^n w_2\|=\infty$. 
How is this possible? I do not have any idea how the norm goes to infinity...Could anyone please help me?
 A: Wlog we can assume $A$ to be in Jordan normal form and the basis is the standard basis. We have
$$ A= I + N$$
where
$$ N= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0& 0 &0 \end{pmatrix}$$
Then $N^3= 0$ and we get $A^n= I + n N + \binom{n}{ 2}N^2$. Thus
$$ A^n e_2 = \begin{pmatrix} n \\ 1 \\ 0 \end{pmatrix} $$
and
$$ A^n e_3 = \begin{pmatrix} \binom{n}{2} \\ n \\ 1\end{pmatrix}. $$
Hence, the norms blow up.
A: More generally, let $\lambda$ be a (multiple) eigenvalue of $A$ and let $w$ be a member of the generalised eigenspace of$~A$ for$~\lambda$ (so $(A-\lambda I)^kw=0$ for some$~k\in\Bbb N$), but not an eigenvector (so $k=1$ will not do: $(A-\lambda I)w\neq0$). Write $N=A-\lambda I$, and fix the minimal $k$ such $N^kw=0$; then $[w,Nw,N^2w,\ldots,N^{k-1}w]$ are linearly independent, so a basis of a subspace, and this subspace is of dimension at least$~2$. We can apply the binomial theorem to $(\lambda I+N)^n$ (since $I$ commutes with $N$) giving
$$A^nw=(\lambda I+N)^n(w)=\sum_{i=0}^{k-1}\binom ni\lambda^{n-i}N^iw,$$
with the expansion being cut off at $i=k$ due to $N^kw=0$. But this is an expansion on the mentioned basis. Clearly the coefficient $\binom ni\lambda^{n-i}$ of this expansion for any $0<i<k$ diverges as $n\to\infty$ if (and only if) $|\lambda|\geq1$, in particular for $\lambda=1$ of the question. In fact these coefficients grow without bound, and so therefore does any norm applied to $A^nw$.
A: The theorem is correct. Without loss of generality and by using Schur decomposition, we can write the most general form of matrix $A$ as$$A=\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}$$Since the eigenspace of $A$ has dimension $1$, and the eigenvector is $v=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$ therefore the following set of equations$$\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$$must set two of $\{v_1,v_2,v_3\}$ equal to zero. Since the set can be reduced to$$av_2+bv_3=0\\cv_3=0$$therefore we can have $v_1\ne0$ and $v_2=v_3=0$. The only case this happens is when$$a,c\ne 0$$. Therefore the only eigenvector of $A$ becomes $$\begin{bmatrix}1\\0\\0\end{bmatrix}$$Now take $w_1=\begin{bmatrix}0\\w_{12}\\w_{13}\end{bmatrix}$ and $w_2=\begin{bmatrix}0\\w_{22}\\w_{23}\end{bmatrix}$ where $w_2\ne kw_1$. One can prove using induction that$$A^nw_1=\begin{bmatrix}naw_{12}+nbw_{13}+{n(n-1)\over 2}acw_{13}\\w_{12}+ncw_{13}\\w_{13}\end{bmatrix}\\A^nw_2=\begin{bmatrix}naw_{22}+nbw_{23}+{n(n-1)\over 2}acw_{23}\\w_{22}+ncw_{23}\\w_{23}\end{bmatrix}$$Since $a,c\ne 0$, it is easy to check that$$||A^nw_1||\to \infty\\||A^nw_2||\to \infty$$$\blacksquare$
P.S.
The general form of $A$ has been chosen upper-triangular since from the Schur decomposition and for any arbitrary vector $v$ we have $$||A^nv||{=||QU^nQ^Hv||\\=||QU^nw||\\=||U^nw||}$$where $w\triangleq Q^Hv$ and $U$ and $Q$ are upper-triangular and unitary respectively.
