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Linear Algebra: Companion notes

Abhay Soman

Section 7.6 Some Computations with Jordan block

  1. Suppose that \(A=(a_{ij})\) be an \(n\times n\) matrix over a field \(F\text{.}\) We describe the collection of elements \(a_{ij}\) with
    1. \(i=j\) the main diagonal of \(A\);
    2. \(i-j=r\) (where \(r\) is a given number) the diagonal lying \(r\) steps from the main diagonal
  2. Let \(A\in M_m(F)\) be the Jordan block (see Definition 7.5.7). Then, \(A=\lambda\cdot I_m+B\) where \(B\) is a matrix with a lower diagonal (diagonal lying \(1\) step from the main diagonal) entries \(1\) and all other entries \(0\text{.}\) We thus have
    \begin{equation*} Be_1^t=e_2^t,\;Be_2^t=e_3^t,\;\ldots,\;Be_{n-1}^t=e_n,\;Be_m^t=0. \end{equation*}
    This shows that
    \begin{equation*} B^re_i^t=e_{i+r}^t\quad\text{for}\;1\leq i\leq m-r\quad\text{and}\quad B^re_i^t=0\quad\text{for}\;i>m-r. \end{equation*}
    Thus, the matrix of \(B^r\) has the diagonal lying \(r\) steps from the main diagonal containing all \(1\)’s, and all other entries are \(0\text{.}\)
  3. Furthermore, \(\lambda I_m\cdot B=B\cdot\lambda I_m\text{.}\) Therefore using (A.1.1), for a polynomial of degree \(n\) in one variable \(f(t)\in F[t]\text{,}\) we get
    \begin{equation} f(A)=f(\lambda)I_m+f^\prime(\lambda)B+\frac{f^{\prime\prime}(\lambda)}{2}B^2+\cdots+\frac{f^{(n)}(\lambda)}{n!}B^n\tag{7.6.1} \end{equation}
    If we put \(g_r=f^{(r)}(\lambda)\big/r!\) then
    \begin{equation} f(A)=\begin{pmatrix}g_0\amp 0\amp 0\amp \cdots\amp 0\amp 0\\g_1\amp g_0\amp 0\amp\cdots\amp 0\amp 0\\g_2\amp g_1\amp g_0\amp\ddots\amp\amp 0\\\vdots\amp\ddots\amp\ddots\amp\ddots\amp\ddots\amp\vdots\\g_{m-2}\amp g_{m-3}\amp\ddots\amp\ddots\amp g_0\amp 0\\g_{m-1}\amp g_{m-2}\amp g_{m-3}\amp\cdots\amp g_1\amp g_0\end{pmatrix}.\tag{7.6.2} \end{equation}
Using the above expression it is relatively easy to compute the powers of a Jordan block. As an exercise compute the \(100\)-th power of \(2\times 2\) Jordan block.
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