How to confirm the $k$-th power of this matrix?

Question

For any $k \in \mathbb N$,

$$\begin{bmatrix}2&-2\\0.5&0\end{bmatrix}^k = \begin{bmatrix}k+1&-2k\\k/2&-(k-1)\end{bmatrix}$$

I found this empirically using numerical computation, but I cannot prove it. I have tried computing the eigendecomposition, but this matrix is not diagonalizable.

Answer 1

Hint

Induction should work better.

Answer 2

Using SymPy:

>>> from sympy import *
>>> A = Matrix([[            2,  -2],
                [Rational(1,2),   0]])

Computing the Jordan normal form,

>>> P, J = A.jordan_form()
>>> P
Matrix([[-5/2, -1/2],
        [-5/4,    1]])
>>> J
Matrix([[1, 1],
        [0, 1]])

Using the binomial theorem, we conclude that the $k$-th power of Jordan matrix $\rm J$ is $\begin{bmatrix} 1 & k\\ 0 & 1\end{bmatrix}$. Thus,

>>> k = Symbol('k', integer=True)
>>> P * Matrix([[1, k],[0, 1]]) * P**-1
Matrix([[k + 1,   -2*k],
        [  k/2, -k + 1]])

However, SymPy is "smart" enough to compute the $k$-th power of $\rm A$:

>>> A**k
Matrix([[k + 1,   -2*k],
        [  k/2, -k + 1]])

Answer 3

Using the Cayley-Hamilton Theorem, the characteristic polynomial is: $$p(\lambda)=\det(\lambda I_2-A)=\begin{vmatrix}\lambda-2&2\\ -0.5&\lambda\end{vmatrix}=\lambda^2-2\lambda+1=0.$$ Hence: $$p(A)=A^2-2A+I=0 \Rightarrow A^2=2A-I.$$ Thus: $$\begin{align}A^2&=2A-I;\\ A^3&=A^2A=(2A-I)A=2A^2-A=2(2A-I)-A=3A-2I;\\ A^4&=A^3A=(3A-2I)A=3A^2-2A=3(2A-I)-2A=4A-3I;\\ \vdots \\ A^k&=kA-(k-1)I=k\begin{pmatrix}2&-2\\0.5&0\end{pmatrix}-(k-1)\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}k+1&-2k\\0.5k&-(k-1)\end{pmatrix}.\\ \end{align}$$

Answer 4

The matrix (call it $A$) may not be diagonalizable, but it does split into the sum of the identity matrix and a nilpotent matrix: $$N = A-I = \begin{bmatrix}1&-2\\\frac12&-1\end{bmatrix}.$$ Since $N$ and $I$ commute you can expand $A^k$ using the binomial theorem: $$A^k = I + kN + \binom k2N^2 + \dots,$$ but since $N^2=0$, all of the terms that involve that and higher powers of $N$ vanish and you have $A^k=I+kN=kA+(1-k)I$.

This method works for any $2\times2$ matrix with a repeated eigenvalue $\lambda$: it will split into $A=\lambda I+N$ with $N^2=0$, and by the binomial formula $$A^k = \lambda^k I+k\lambda^{k-1}N = k\lambda^{k-1}A + (1 - k)\lambda^k I.$$