MHBCalculate A^17: Powers of Matrices

[Petrus]

Hello MHB,
Calculate \(\displaystyle A^{17}\) where

.

Progress,
I have multiplicate without adding them together to see a pattern and I can se at \(\displaystyle A^{17}\) on that matrice where it's 6's it will be \(\displaystyle 6^{17}\) and rest I can't se any pattern those riight side of the triangle, cause the left will be zero

Regards,

 

[caffeinemachine]

Hello MHB,
Calculate \(\displaystyle A^{17}\) where

.

Progress,
I have multiplicate without adding them together to see a pattern and I can se at \(\displaystyle A^{17}\) on that matrice where it's 6's it will be \(\displaystyle 6^{17}\) and rest I can't se any pattern those riight side of the triangle, cause the left will be zero

Regards,

Cayley Hamilton theorem might be useful. Cayley

 

[tkhunny]

A little exploration shows:

If $$A = \left( \begin{array}{ccc} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & a \end{array} \right)$$

Then $$A^{17} = \left( \begin{array}{ccc} a^{17} & 17a^{16} & 8\cdot 17\cdot a^{15} \\ 0 & a^{17} & 17a^{16} \\ 0 & 0 & a^{17} \end{array} \right)$$

Upper triangular with only '1' off the diagonal? That HAS to be rather tractable.

 

[Fernando Revilla]

In general, if we have a Jordan block $J(\lambda)$ of order $n$ we can express $$J(\lambda)=\begin{bmatrix} \lambda & 1 & 0 &\ldots & 0 & 0 & 0\\ 0 & \lambda & 1 &\ldots & 0&0&0 \\0 & 0 & \lambda &\ldots & 0&0&0 \\\vdots&&&&&&\vdots \\ 0 &0 & 0 &\ldots & \lambda & 1&0\\0 &0 &0 &\ldots &0&\lambda & 1\\0 & 0 &0&\ldots & 0&0&\lambda\end{bmatrix}=\lambda I+N$$ where $N$ is nilpotent of order $n$. As $(\lambda I_n)N=N(\lambda I_n)$, we can apply the binomial theorem $$J(\lambda)^m=(\lambda I+N)^m=(\lambda I)^{m}+\binom{m}{1}(\lambda I)^{m-1}N+\binom{m}{2}(\lambda I)^{m-2}N^2+\ldots\\=\lambda^mI+m\lambda^{m-1} N+\frac{m(m-1)}{2}\lambda^{m-2}N^2+\ldots$$ If $n=3$, $N$ is nilpotent of order $2$: $$N=\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix},N^2=\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix},N^3=0$$ $$J(\lambda)^m=\lambda^mI+m\lambda^{m-1} N+\frac{m(m-1)}{2}\lambda^{m-2}N^2$$ $$\begin{bmatrix}{\lambda}&{1}&{0}\\{0}&{\lambda}&{1}\\{0}&{0}&{\lambda}\end{bmatrix}^m=\lambda^m \begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}+m\lambda^{m-1}\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix}+\frac{m(m-1)}{2}\lambda^{m-2}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}$$ $$=\begin{bmatrix}{\lambda^m}&{m\lambda^{m-1}}&{\frac{m(m-1)}{2}}\lambda^{m-2}\\{0}&{\lambda^m}&{m\lambda^{m-1}}\\{0}&{0}&{\lambda^m}\end{bmatrix}$$

 

[blue_raver22]

I would like to provide a more detailed explanation of how to calculate A^{17}. First, let's define A as a matrix with elements a_{ij} where i and j represent the row and column indices, respectively. In order to calculate A^{17}, we can use the power of matrices formula:

A^{17} = (A^2)^8 * A

This formula utilizes the fact that when multiplying matrices, the exponent of the resulting matrix is equal to the sum of the exponents of the original matrices. Therefore, we can calculate A^{17} by first calculating A^2, then raising it to the 8th power, and finally multiplying it by A.

Now, let's focus on calculating A^2. This can be done by multiplying A with itself:

A^2 = A * A

Since A is a 3x3 matrix, the resulting matrix will also be a 3x3 matrix. The elements of the resulting matrix can be calculated by using the matrix multiplication rule, where the element at position (i,j) is equal to the sum of products of the elements in the ith row of A and the jth column of A.

Once we have calculated A^2, we can raise it to the 8th power by multiplying it with itself 8 times. This will result in a new 3x3 matrix, which we can then multiply with A to get the final result of A^{17}.

In summary, calculating A^{17} requires a series of matrix multiplications and may not have a clear pattern like the one observed for A^{6}. As a scientist, it is important to carefully follow mathematical rules and formulas to accurately calculate results.