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If$$ A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \\ \end{pmatrix} $$ prove that $$A^k = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \\ \end{pmatrix}$$

Now the first method I used was using assuming an equation $$x^k = f(x)(Ax^2 + Bx +C) + (px + q)........(i)$$

where $ Ax^2 + Bx + C = 0 $ is characteristic equation and $x$ is eigen value.

After solving the characteristic equation I would get two eigen value $x_1$ & $ x_2$

Putting those two eigen values in $(i)$ will give me two equations and I will be able to find $p$ and $q$.

Replacing the value of $p$ and $q$ in $(i)$ and replacing $x$ with $A$ (Cayley Hamilton Theorem) I should be able to get the value of $A^k$

but the problem is that $A$ has 2 identical eigen value ($i.e. 1$) so $p$ and $q$ will have infinite solutions. Thus I cannot apply this method.

I found this other method of finding $n^th$ power of matrices online which uses diagonalization of matrices

but as $A$ has only 1 eigen value therefore it also has only 1 eigen vector. Thus $A$ cannot be diagonalized.

I can just simply multiply $A$ 2 or 3 times and derive the general formula but I am looking for another way.

Thank you for taking your time and reading this far.

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Let's prove this by induction. If $k=1$, the result is clear. Now, if $$ A^k = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \\ \end{pmatrix}, $$therefore $$ A^{k+1} =A \times A^k = \begin{pmatrix} 3 & -4 \\ 1 & -1 \\ \end{pmatrix} \times \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \\ \end{pmatrix} =\begin{pmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1) \\ \end{pmatrix} $$ after simplifying.

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The trace of $A$ is $2$ and the determinant is $1$. So, the characteristic polynomial of $A$ is $x^2-2x+1$.

By Cayley–Hamilton, we have $A^2=2A-I$ and so $A^3=2A^2-A=2(2A-I)-A=3A-2I$.

By induction, $A^k=kA - (k-1)I$, that is $$ A^k = \begin{pmatrix} 3k & -4k \\ k & -k \end{pmatrix} - \begin{pmatrix} k-1 & 0 \\ 0 & k-1 \end{pmatrix} = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \end{pmatrix} $$

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If I understand your method correctly, you can use it, but you have to modify it a bit to work with repeated eigenvalues. We write $A^k=pA+qI$ and then find the unknown coefficients $p$ and $q$ by substituting the eigenvalues of $A$ for $A$ and solving the resulting system of equations. The problem that you’ve run into here is that with a repeated eigenvalue, you don’t have enough independent equations to obtain a unique solution for the system. This is easily remedied, though: you can generate additional independent equations via differentiation. So, in this case the system of equations that you need to solve is $$p\lambda+q = \lambda^k \\ p = k\lambda^{k-1},$$ from which $q=(1-k)\lambda^k$ and so $$A^k = k\lambda^{k-1}A+(1-k)\lambda^kI = \lambda^kI+k\lambda^{k-1}(A-\lambda I).$$

Another approach is to set $N=A-\lambda I$ and expand $(\lambda I+N)^k$ using the Binomial theorem, noting that $\lambda I$ and $N$ commute and that $N^2=0$.

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