# Finding n-th power of a 2*2 matrix with 2 identical eigen values

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.

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.
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 $$A^k = \begin{pmatrix} 3k & -4k \\ k & -k \end{pmatrix} - \begin{pmatrix} k-1 & 0 \\ 0 & k-1 \end{pmatrix} = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \end{pmatrix}$$
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 $$N^2=0$$.