Find Matrix $A^{50}$?

Question

Find the matrix $A^{50}$ given

$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$

I was practicing some questions for my exam and I found questions of this form in a previous year's paper.

I don't know how to do such questions.

Please assist over this question.

Thank You

Answer 1

This is for the updated version of question where $A = \begin{bmatrix}2 & -1\\0 & 1\end{bmatrix}$. For the original version of $A$, the derivation is similar.

The characteristic polynomial for matrix $A$ is

$$\chi_A(\lambda) = \det\begin{bmatrix}\lambda-2 & 1\\0 & \lambda - 1\end{bmatrix} = (\lambda - 2)(\lambda - 1) = \lambda^2 - 3\lambda + 2$$

By Cayley Hamilton theorem, we have

$$A^2 - 3A + 2I_2 = 0$$

If we divide the polynomial $x^{50}$ by $x^2 - 3x + 2$ with long division, we know there are polynomial $p(x)$ and coefficients $\alpha, \beta$ such that

$$x^{50} = p(x)(x^2 - 3x + 2) + \alpha x + \beta$$

To determine $\alpha, \beta$, substitute $x$ by $1$ and $2$ in above expression. We get

$$\begin{cases} 1 &= \alpha + \beta\\ 2^{50} &= 2\alpha + \beta \end{cases} \quad\implies\quad \begin{cases} \alpha &= 2^{50} - 1\\ \beta &= 2 - 2^{50} \end{cases}$$ Form this, we get

$$\require{cancel} \begin{align} A^{50} &= p(A)\color{red}{\cancelto{0}{\color{gray}{(A^2 - 3A + 2I_2)}}} + \alpha A + \beta = \alpha A + \beta\\ &= (2^{50}-1)\begin{bmatrix}2 & -1\\0 & 1\end{bmatrix} +(2 - 2^{50})\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} = \begin{bmatrix}2^{50} & 1 - 2^{50}\\0 & 1\end{bmatrix} \end{align} $$

Answer 2

Hint: Diagonalize the matrix you have been given and then take powers.

Answer 3

You can also compute it with the following method. First note that $50 = 32 + 16 + 2$. Let $$A=\left(\begin{array}{cc}2&0\\2&1\end{array}\right).$$

Then $$A^{50} = A^{32} A^{16} A^2$$.

We can compute $A^{2^n}$ easier than $A^{50}$,

$$A^2 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)$$

$$A^4 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)\left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)= \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right)$$

$$A^8 = \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)$$

$$A^{16} = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)\left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)=\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)$$

and

$$A^{32} = \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)=\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right).$$

Finally $$A^{50} =\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right) \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right) \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right).$$

This isn't the best way to go about it, but it is a lot easier than multiplying $A$ to itself 50 times. You should probably persue @science's method to be honest.

Answer 4

Hint: Find a matrix $P$ such that $P^{-1}AP=D$ where $D=diag(2,1)$. Then $P^{-1}A^{50}P=D^{50}=diag(2^{50},1).$

Answer 5

To diagonalize a matrix, you first need to find the eigenvalues, solve:

$$\det(A - \lambda I ) = 0$$ $$\det \begin{bmatrix} 2 - \lambda & 0 \\ 2 & 1 - \lambda \end{bmatrix} = 0$$

$$(1 - L)(2 - L) = 0$$ $$L \in \{1, 2\}$$

So the diagonal values are $1$ and $2$.

Then you need to find the eigenvectors using nullspace basis:

$$\begin{align} V_1 &= {\rm nullspace}(A - 1\times I) \\ & = {\rm nullspace}\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{align}$$
and $$\begin{align} V_2 &= {\rm nullspace}(A - 2\times I) \\ & = {\rm nullspace}\begin{bmatrix} 0 & 0 \\ 2 & -1 \end{bmatrix} \\ & = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{align}$$

So with the eigenvalues and eigenvectors you can contruct:

$$\begin{align} A^{50} &= \left([V_1 ~ V_2]A [V_1~V_2]^{-1}\right)^{50} \\ & = [V_1 ~ V_2]A^{50} [V_1~V_2]^{-1} \\ & = \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1^{50} & 0 \\ 0 & 2^{50} \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 2^{50} & 0 \\ 2^{51} - 2 & 1 \end{bmatrix} \end{align}$$

Answer 6

An unorthodoxical but quick solution.

Let us compute the powers by recurrence:

$$ \begin{bmatrix} a_{n+1} & b_{n+1} \\ c_{n+1} & d_{n+1} \end{bmatrix}= \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a_n & b_n \\ c_n & d_n \end{bmatrix} =\begin{bmatrix} 2a_n-c_n & 2b_n-d_n \\ c_n & d_n \end{bmatrix} ,$$ with the initial values $$\begin{bmatrix} a_0 & b_0 \\ c_0 & d_0 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} .$$ Thanks to invariance, we immediately have $c_n=0$ and $d_n=1$. Then solving simple recurrences $$a_{n+1}=2a_{n}=2^{n+1},$$ $$b_{n+1}=2b_n-1=1-2^{n+1},$$ and $$\begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}^{50}= \begin{bmatrix} 2^{50} & 1-2^{50} \\ 0 & 1 \end{bmatrix} .$$

Answer 7

Look at the first few powers: $$\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)$$ $$A^2=\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)=\left(\begin{matrix}4&-3\\0&1 \end{matrix} \right)$$ $$A^3=\left(\begin{matrix}8&-7\\0&1 \end{matrix} \right)$$ This suggests that $$A^n=\left(\begin{matrix}2^n&1-2^n\\0&1 \end{matrix} \right)$$ Which can be proved by induction.

Answer 8

And yet another approach: Splitting $A$ into a diagonal matrix $D$ and a pertubation matrix $P$.

We look at the first powers of $P$:

$$ P^0 = \left( \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \right) \quad P^1 = \left( \begin{matrix} 0 & -1 \\ 0 & 0 \end{matrix} \right) \quad P^2 = \left( \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right) $$ and note that quadratic powers and higher of $P$ vanish. Then $$ DP = 2 P \quad PD = P $$ Thus \begin{align} A^n &= (D + P)^n \\ &= (D^2 + DP + PD + P^2) (D+P)^{n-2} \\ &= (D^2 + (1+2^1)P) (D+P)^{n-2} \\ &= (D^3 + (1+2^1)PD + D^2P + 3P^2)(D+P)^{n-3} \\ &= (D^3 + (1+2^1)P + 2^2 P)(D+P)^{n-3} \\ &= (D^3 + (1+2^1+2^2) P)(D+P)^{n-3} \\ &= D^n + (1+2^1+\cdots+2^{n-1})P \\ &= D^n + (2^n-1)P \end{align}

This gives $$ A^n = \left( \begin{matrix} 2^n & 0 \\ 0 & 1^n \end{matrix} \right) + \left( \begin{matrix} 0 & 1-2^n \\ 0 & 0 \end{matrix} \right) = \left( \begin{matrix} 2^n & 1-2^n \\ 0 & 1 \end{matrix} \right) $$