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Given that $$A=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$$ $$X=\begin{bmatrix}2&1\\ 3&2\end{bmatrix}$$

I computed that $$X^{-1} A X = \begin{bmatrix}7&4\\ \:\:\:\:-9&-5\end{bmatrix}=Y$$

Now the usual way would be to find the eigenvalues and corresponding eigenvectors and then calculate the nth degree for this matrice, however it turns out that this matrix has only one eigenvalues and one eigenvector so I can't use that method.

Instead I thought of using the Cayley-Hammilton theorem and after finding the characteristic polynomial we have:

$Y^2=2Y-E$ which ultimately yields $Y^n = nY-(n-1)E$ is my approach correct?

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    $$(X^{-1} A X)^n = X^{-1} A X X^{-1} A X \cdots X^{-1} A X = X^{-1} A^n X$$ Now note that $A$ is a Jordan block. – 2017-01-09
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    indeed it's the case thanks, but my longer and "naive" approach was also correct ?(I need that confirmation because of other problems where A isn't a Jordan block). – 2017-01-09
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    Yes your approach works as well. To compute $Y^n$ directly, note that $A^n=\begin{pmatrix}1&n\\0&1\end{pmatrix}$ hence $A^n=nA-(n-1)E$ and $Y^n=X^{-1}A^nX=X^{-1}(nA-(n-1)E)X=nX^{-1}AX-(n-1)X^{-1}EX=nY-(n-1)E$. – 2017-01-09

3 Answers 3

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Your approach is correct. As you said, you know that $$(Y-E)^2=0$$ is the zero matrix. In this way, to compute $Y^n$ you should calculate the remainder of dividing the polynomial $y^n$ by $(y-1)^2$ and evaluate it on $Y$ (as the divisor times the quotient will vanish when you evaluate $Y$ on it and so only remains the remainder evaluated at $Y$). In this particular case, write $$y^n=((y-1)+1)^n=\sum_{k=0}^n\binom{n}{k}(y-1)^k$$ and then it is clear that the remainder is $$n(y-1)+1=ny+(n-1)$$ and so $$Y^n=nY+(n-1)E$$ as you have shown.

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Hint:

first prove, by induction, that $$ A^n=\begin{bmatrix} 1&n\\0&1 \end{bmatrix} $$

Then use the fact (easy to prove): $$ (X^{-1}AX)^n=X^{-1}A^nX $$

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It is not necessary to compute $X^{-1}AX$: $$\left(X^{-1}AX\right)^n=\left(X^{-1}AX\right)\left(X^{-1}AX\right)\ldots \left(X^{-1}AX\right)=X^{-1}A^nX.$$ Now, $$A^n=\left(I+\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}\right)^n\underbrace{=}_{\text{Newton}}I^n+\displaystyle\binom{n}{1}I^{n-1}\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}+\displaystyle\binom{n}{1}I^{n-2}\underbrace{\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}^2}_{\text{Null matrix}}$$ $$+\mathbf{0}+\cdots \mathbf{0}=\begin{bmatrix}{1}&{n}\\{0}&{1}\end{bmatrix}.$$