I want to compute the Matrix $A^{19}$ where
$$A = \begin{pmatrix} 6&1&0\\0&6&1\\0&0&6\end{pmatrix}$$
Since the only eigenvalue of the matrix is 6, it's not diagonalizable. How do you proceed to calculate the exponent matrix of A then?
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 26 '21 at 10:57
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1$A$ is a Jordan block. See https://math.stackexchange.com/questions/326688/why-does-the-n-th-power-of-a-jordan-matrix-involve-the-binomial-coefficient – Jean-Claude Arbaut Aug 26 '21 at 11:00
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Try to find the pattern of $n$ powers with the generic $$\begin{pmatrix} \lambda & 1 & 0 \ 0 & \lambda & 1 \ 0 & 0 & \lambda \end{pmatrix}$$ – Ninad Munshi Aug 26 '21 at 11:00
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Let$$B=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$$Then $A=6\operatorname{Id}_3+B$. Besides, $6\operatorname{Id}_3$ and $B$ commute, and therefore\begin{align}A^{19}&=(6\operatorname{Id}_3+B)^{19}\\&=6^{19}\operatorname{Id}_3+19\times6^{18}B+\binom{19}26^{17}B^2\\&=\begin{bmatrix}6^{19}&19\times 6^{18}&\binom{19}26^{17}\\0&6^{19}&19\times6^{18}\\0&0&6^{19}\end{bmatrix},\end{align}since $B^n=0$ when $n>2$.
José Carlos Santos
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3OP is not looking for $\exp(A)$ but $A^{19}$ (I agree the title is confusing) – Ninad Munshi Aug 26 '21 at 11:02
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Hi, thank you for your answer! I am not able to understand how (19 2)6^17*B^2 happened, why is that? – OLGJ Aug 26 '21 at 19:53
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To be more precise:$$(6\operatorname{Id}_3+B)^{19}=\binom{19}06^3\operatorname{Id}_3^{,19}+6^{18}\binom{19}1\operatorname{Id}_3^{,18}B+6^{17}\binom{19}2\operatorname{Id}_3^{,17}B^2+\cdots$$Besides, $\operatorname{Id}_3^{,n}=\operatorname{Id}_3$ for each $n$. – José Carlos Santos Aug 26 '21 at 20:29
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$ A = \begin{bmatrix} 6 && 1 && 0 \\ 0 && 6 && 1 \\ 0 && 0 && 6 \end{bmatrix} $
We can express $A^n$ as follows:
$A^n = \alpha_0 I + \alpha_1 A + \alpha_2 A^2 $
The above equation is satisfied by $\lambda = 6$, i.e.
$\lambda^n = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda^2 $
And since it is a repeated eigenvalue, we use derivatives of this last equation to obtain independent equations in the three unknowns $\alpha_0, \alpha_1, \alpha_2$, so we have
$ n \lambda^{n-1} = \alpha_1 + 2 \alpha_2 \lambda $
and
$ n(n-1) \lambda^{n-2} = 2 \alpha_2 $
Solving the above three equations for $\alpha_0, \alpha_1, \alpha_2$, we can evaluate the matrix $A^n$.
We have,
$\alpha_2 = (19)(9) (6)^{17} $
$\alpha_1 = 19 (6)^{18} - (19)(18) 6^{18} = - 323 (6)^{18}$
$\alpha_0 = (6)^{19} + (323) 6^{19} - (19)(9) 6^{19} = 153 (6)^{19}$
and we have
$A^2 = \begin{bmatrix} 6^2 && 12 && 2 \\ 0 && 6^2 && 12 \\ 0 && 0 && 6^2 \end{bmatrix} $
Putting it all together we can find the entries of $A^{19}$
$A^{19} = 153 (6)^{19} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - 323 (6)^{18} \begin{bmatrix} 6 && 1 && 0 \\ 0 && 6 && 1 \\ 0 && 0 && 6 \end{bmatrix} + 171 (6)^{17} \begin{bmatrix} 6^2 && 12 && 2 \\ 0 && 6^2 && 12 \\ 0 && 0 && 6^2 \end{bmatrix} $
Which reduces to,
$A^{19} = \begin{bmatrix} 6^{19} && 19(6)^{18} && 57(6)^{18} \\ 0 && 6^{19} && 19 (6)^{18} \\ 0 && 0 && 6^{19} \end{bmatrix} $