Bloque de Jordan

En los siguiente ejercicios usamos el concepto de bloque de Jordan.

RESUMEN TEÓRICO
    Enunciado
  1. Comprobar que la siguiente matriz es nilpotente de orden $4$ $$A=\begin{bmatrix}{0}&{1}&{0}& 0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\0&0&0&0\end{bmatrix}.$$
  2. Hallar $A^n$ siendo $A=\begin{bmatrix}{3}&{1}\\{0}&{3}\end{bmatrix}.$
  3. Calcular $A^n,$ siendo $A=\begin{bmatrix}{3}&{1}&{0}& 0&0
    \\{0}&{3}&{0} & 0 & 0
    \\{0}&{0}&{-1}& 1& 0
    \\{0}&{0}&{0} & -1 & 1
    \\{0}&{0}&{0} & 0 &-1\end{bmatrix}.$
    Solución
  1. Tenemos: $$A^2=\begin{bmatrix}{0}&{1}&{0}& 0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\0&0&0&0\end{bmatrix}\begin{bmatrix}{0}&{1}&{0}& 0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\0&0&0&0\end{bmatrix}=\begin{bmatrix}{0}&{0}&{1}& 0\\{0}&{0}&{0}&1\\{0}&{0}&{0}&0\\0&0&0&0\end{bmatrix},$$ $$A^3=\begin{bmatrix}{0}&{0}&{1}& 0\\{0}&{0}&{0}&1\\{0}&{0}&{0}&0\\0&0&0&0\end{bmatrix}\begin{bmatrix}{0}&{1}&{0}& 0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\0&0&0&0\end{bmatrix}=\begin{bmatrix}{0}&{0}&{0}& 1\\{0}&{0}&{0}&0\\{0}&{0}&{0}&0\\0&0&0&0\end{bmatrix},$$ $$A^4=\begin{bmatrix}{0}&{0}&{0}& 1\\{0}&{0}&{0}&0\\{0}&{0}&{0}&0\\0&0&0&0\end{bmatrix}\begin{bmatrix}{0}&{1}&{0}& 0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\0&0&0&0\end{bmatrix}=\begin{bmatrix}{0}&{0}&{0}& 0\\{0}&{0}&{0}&0\\{0}&{0}&{0}&0\\0&0&0&0\end{bmatrix}.$$ Es decir, $A$ es nilpotente de orden $4.$
  2. Tenemos $A=3I+N$ con $I$ la matriz identidad y $N=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}.$ Entonces:$$N^2=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}=\begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix},\;N^3=N^4=\ldots=\begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}$$Aplicando la fórmula del binomio de Newton:$$\begin{aligned}A^n&=\left(3I+N\right)^n=\binom{n}{0}(3I)^n+\binom{n}{1}(3I)^{n-1}N+\binom{n}{2}(3I)^{n-2}N^2+\cdots\\
    &=3^nI+n3^{n-1}\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}+\begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}+\cdots+\begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}\\
    &=3^n\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}+n3^{n-1}\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}=\begin{bmatrix}{3^n}&{n3^{n-1}}\\{0}&{3^n}\end{bmatrix}.\end{aligned}$$
  3. La matriz $A$ es de la forma $$A=\begin{bmatrix}{A_1}&{0}\\{0}&{A_2}\end{bmatrix}\text{ con }A_1=\begin{bmatrix}{3}&{1}\\{0}&{3}\end{bmatrix},\;A_2=\begin{bmatrix}{-1}&{1}& 0\\{0}&{-1}& 1\\0&0&-1\end{bmatrix},$$ por tanto la matriz $A^n$ es $A^n=\begin{bmatrix}{A_1^n}&{1}\\{0}&{A_2^n}\end{bmatrix}.$ La matriz $A_1^n$ se calculó en el apartado anterior: $$A_1^n=\begin{bmatrix}{3^n}&{n3^{n-1}}\\{0}&{3^n}\end{bmatrix}.$$ Usamos la fórmula del binomio de Newton para hallar $A_2^n.$ Denotando $$N=\begin{bmatrix}{0}&{1}& 0\\{0}&{0}& 1\\0&0&0\end{bmatrix}:$$ $$\begin{aligned}A_2^n&=\left(-I+N\right)^n=\binom{n}{0}(-I)^n+\binom{n}{1}(-I)^{n-1}N+\binom{n}{2}(-I)^{n-2}N^2+\cdots\\
    &=(-1)^nI+n(-1)^{n-1}\begin{bmatrix}{0}&{1}&0\\{0}&{0}& 1\\0 & 0 & 0\end{bmatrix}+\frac{n(n-1)}{2}\begin{bmatrix}{0}&{0}&1\\{0}&{0}& 0\\0 & 0 & 0\end{bmatrix}
    \\&+\begin{bmatrix}{0}&{0}&0\\{0}&{0}& 0\\0 & 0 & 0\end{bmatrix}+\cdots+\begin{bmatrix}{0}&{0}&0\\{0}&{0}& 0\\0 & 0 & 0\end{bmatrix}\\
    &=\begin{bmatrix}{(-1)^n}&{(-1)^{n-1}n}&(-1)^{n-2}n(n-1)/2\\{0}&{(-1)^n}& (-1)^{n-1}n\\0 & 0 & (-1)^n\end{bmatrix}\end{aligned}$$ Queda por tanto $$A^n=\begin{bmatrix}3^n & n3^{n-1} & 0 & 0 & 0\\0 & 3^n & 0 &0 & 0\\0 & 0 &{(-1)^n}&{(-1)^{n-1}n}&(-1)^{n-2}n(n-1)/2\\0 & 0 & {0}&{(-1)^n}& (-1)^{n-1}n\\0 & 0 &0 & 0 & (-1)^n\end{bmatrix}.$$
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