Suppose that $T$ is a linear operator on $C^8$ and $\beta = \{v_1, v_2, \cdots, v_8\}$ is an ordered basis for $C^8$ such that:
$J = [T]_\beta = \left( \begin{matrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0&0 \\ 0&0&0&0&3&1&0&0 \\ 0&0&0&0&0&3&0&0 \\ 0 & 0&0&0&0&0&0&1 \\ 0&0&0&0&0&0&0&0 \end{matrix} \right)$
is a Jordan canonical form of T.
How can I immediately recognize that $T(v_2) = v_1 + 2v_2$? How can I immediately recognize that $(T - 2I)^3(v_i) = 0$?
These are things my book just throws out there with no explanation and I don't understand how to come to these conclusions just by looking at the matrix.
Thank you.
You can think of the i-th column as being the coefficients for the linear combination which gives you $T(v_i)$ in terms of the basis $\{v_1,...,v_8\}$.
So if we look at the second column we can see it indicates that the image of $T(v_2)=v_1+2v_2$.
As another example, we can see the last column is nothing but zeroes except for the second to last entry. This indicates that if we were to write out $T(v_8)$ in terms of the basis $\beta$, i.e. $T(v_8)=\sum_{i=1}^8a_iv_i$ that the only non zero coefficient is $a_7$ which in the case $a_7=1$, and so we can conclude that $T(v_8)=v_7$.
Now for the second part of your question, we use the same reasoning as we did for the first part, but we first need to subtract a 2 from every element in the diagonal. So based on what I previously wrote, what can we infer about this new matrix based on the entries of the matrix?