Jordan Normal Form - Number of Ones on Superdiagonal

Question

I was reading notes on Jordan Normal Form and it says that for a given matrix $A$, the number of ones on the super-diagonal of its associated Jordan matrix is equal to $n-d$, however they seem to assume an implied meaning of $n$ and $d$ which I can't seem to figure out. If you scroll to the bottom, where they mention it, they use $d$ once saying that the Jordan Matrix is of the form $$\left(\begin{array}{cccc} J_1 & 0 & \cdots & 0 \\ 0 & J_2 & \cdots & 0 \\ \vdots & \vdots & \ddots&\vdots \\ 0 & 0 & \cdots & J_d \\ \end{array}\right)$$ appearing to refer to the number of Jordan blocks. They also use $n$ on the bullet above referring to the number of distinct eigenvalues of $A$. Is this interpretation of $n$ and $d$ most likely what they meant (i.e. does $n-d$ give the number of ones on the superdiagonal)? If so can anyone explain why?

Answer 1

The number $n$ in $n-d$ is the total size of the matrix, that is the the sum of the sizes of $J_1,\ldots,J_d$. If $J_r$ is of size $k_r$, then $$k_1+\cdots+k_d=n.$$ Each block has $k_r-1$ ones, so the total number of ones is $$ (k_1-1)+\cdots+(k_d-1)=(k_1+\cdots+k_d)-d=n-d. $$