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\)
Section 32.5 Examples
Example 32.5.1 . Putting a nilpotent matrix into elementary nilpotent form.
\begin{equation*}
A = \begin{abmatrix}{rrrr}
1 \amp 2 \amp 0 \amp -1 \\
3 \amp 7 \amp -1 \amp -4 \\
-1 \amp 9 \amp -2 \amp -4 \\
6 \amp 10 \amp -1 \amp -6
\end{abmatrix}\text{.}
\end{equation*}
Compute the powers of \(A\text{:}\)
\begin{align*}
A^2 \amp = \begin{abmatrix}{ccrr}
1 \amp 6 \amp -1 \amp -3 \\
1 \amp 6 \amp -1 \amp -3 \\
4 \amp 3 \amp -1 \amp -3 \\
1 \amp 13 \amp -2 \amp -6
\end{abmatrix} \text{,}
\amp
A^3 \amp = \begin{abmatrix}{rrrr}
2 \amp 5 \amp -1 \amp -3 \\
2 \amp 5 \amp -1 \amp -3 \\
-4 \amp -10 \amp 2 \amp 6 \\
6 \amp 15 \amp -3 \amp -9
\end{abmatrix} \text{,}
\amp
A^4 \amp = \zerovec \text{.}
\end{align*}
Since every column of \(A^3\) is nonzero, we can choose \(\uvec{v}\) to be any of the standard basis vectors. Let’s choose \(\uvec{v} = \uvec{e}_1\text{.}\) Then, using
\begin{align*}
\uvec{p}_1 \amp= \uvec{v} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \text{,} \amp
\uvec{p}_2 \amp= A\uvec{v} = \begin{abmatrix}{r} 1 \\ 3 \\ -1 \\ 6 \end{abmatrix} \text{,} \amp
\uvec{p}_3 \amp= A^2 \uvec{v} = \begin{bmatrix} 1 \\ 1 \\ 4 \\ 1 \end{bmatrix} \text{,} \amp
\uvec{p}_4 \amp= A^3 \uvec{v} = \begin{abmatrix}{r} 2 \\ 2 \\ -4 \\ 6 \end{abmatrix} \text{,}
\end{align*}
we get
\begin{equation*}
P =
\begin{abmatrix}{crcr}
1 \amp 1 \amp 1 \amp 2 \\
0 \amp 3 \amp 1 \amp 2 \\
0 \amp -1 \amp 4 \amp -4 \\
0 \amp 6 \amp 1 \amp 6
\end{abmatrix}\text{.}
\end{equation*}
Because of our choice of \(\uvec{v} = \uvec{e}_1\text{,}\) there wasn’t any need to actually compute the products \(A^j \uvec{v}\) to obtain the columns of \(P\text{,}\) as \(A^j \uvec{e}_1\) is just the first column of \(A^j\text{.}\)
There also isn’t any need to compute \(\inv{P} A P\text{,}\) as we know our procedure will result in elementary nilpotent form
\begin{equation*}
\inv{P}AP =
\begin{bmatrix}
0 \amp 0 \amp 0 \amp 0 \\
1 \amp 0 \amp 0 \amp 0 \\
0 \amp 1 \amp 0 \amp 0 \\
0 \amp 0 \amp 1 \amp 0
\end{bmatrix}\text{.}
\end{equation*}
