Jordan block and Jordan chain question? I don't understand why it's apparently 'clear' that the matrix of $T$ with respect to the basis $v_1, \dots, v_n$ is a Jordan block of degree $n$ if and only if $v_1, \dots ,v_n$ is a Jordan chain for $A$. Can anyone explain this? Thanks! 
 A: Writing $T(v_i)$ as a linear combination gives you the column of the matrix of $T$ with respect to this Basis. But for a Jordan chain
$$
T(v_1)=\lambda v_1 \quad T(v_2)=v_1 + \lambda v_2 \quad T(v_3)=v_2+\lambda v_3 \dots
$$
A: Suppose that $F$ is a field, $V$ is an $n$-dimensional vector space, and $T:V\to V$ is a $F$-linear map. Suppose that $v_1,\ldots, v_n$ is a basis of $V$ such that $(T-\lambda I)v_k=v_{k-1}$ for $1\leq k\leq n$, where $v_0=0$ is the zero vector. Then:
$$(T-\lambda I)^k v_n=v_{n-k},$$
for all $1\leq k \leq n$ and, in particular,
$$(T-\lambda I)v_1 = (T-\lambda I)(T-\lambda I)^{n-1}v_n = (T-\lambda I)^n v_n=0.$$
Therefore, $T(v_1)=\lambda v_1$ and $v_1$ is an eigenvector with eigenvalue $\lambda$. Moreover, $(T-\lambda I)v_{k}=v_{k-1}$ and this implies $Tv_k = \lambda v_k + v_{k-1}$, for each $1\leq k\leq n$. Therefore, the matrix for $T$ in terms of the basis $\{v_1,\ldots,v_n\}$ of $V$ has the form:
$$\left(\begin{array}[ccccccc]

& \lambda & 1 & 0& 0 & \cdots & 0& 0 \\
 0 & \lambda & 1 & 0 & \cdots &0 & 0\\
0 & 0 & \lambda & 1 &  \cdots &0 & 0\\
\vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\
0 & 0 & 0 & 0 & \cdots &1 & 0\\
0 & 0 & 0 & 0 & \cdots &\lambda & 1\\
0 & 0 & 0 & 0 & \cdots & 0 & \lambda\\

\end{array}\right),$$
which is clearly a Jordan block.
Conversely, if the matrix of $T$ is a Jordan block as above with respect to a basis $\{v_1,\ldots, v_n\}$, then $T(v_1)=\lambda v_1$ and $T(v_k)=\lambda v_k + v_{k-1}$, or equivalently, $(T-\lambda I)v_1=0$ and $(T-\lambda I)v_k=v_{k-1}$, so the vectors in the basis form a Jordan chain.
