Find $2^A$ with $A$ is a matrix

Question

Let

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

How to find $2^A$ ?

I find out that $A^2=I$ so it would be simple if they ask me how to find a power of $A$, but not. So could you help me?

Answer 1

\begin{align}2^A&=e^{\log(2)A}\\&=\operatorname{Id}+\log(2)A+\frac{\log(2)^2A^2}{2!}+\frac{\log(2)^3A^3}{3!}+\cdots\\&=\operatorname{Id}+\log(2)A+\frac{\log(2)^2\operatorname{Id}}{2!}+\frac{\log(2)^3A}{3!}+\cdots\\&=\left(1+\frac{\log^2(2)}{2!}+\frac{\log^4(2)}{4!}+\cdots\right)\operatorname{Id}+\left(\log(2)+\frac{\log^3(2)}{3!}+\cdots\right)A\end{align}

Answer 2

Hint. One has $$ e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots+\frac{A^n}{n!}+\cdots $$ then use that $$ A^2=I,\,A^3=A,\,\cdots. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\textsf{A}^{1} = \textsf{A}\,,\ \textsf{A}^{2} = \textsf{1} \implies \expo{x\textsf{A}} = \mrm{p}\pars{x}\textsf{1} + \mrm{q}\pars{x}\textsf{A}}$

with $\ds{\mrm{p}\pars{0} = 1\,,\mrm{q}\pars{0} = 0}$.

\begin{align} &\left.\begin{array}{rcl} \ds{A\expo{x\textsf{A}}} & \ds{=} & \ds{\mrm{p}\pars{x}'\textsf{1} + \mrm{q}'\pars{x}\textsf{A}} \\ \ds{A\expo{x\textsf{A}}} & \ds{=} & \ds{\mrm{p}\pars{x}\textsf{A} + \mrm{q}\pars{x}\textsf{1}} \end{array}\right\} \implies \left\{\begin{array}{rcl} \ds{\mrm{p}'\pars{x}} & \ds{=} & \ds{\mrm{q}\pars{x}} \\ \ds{\mrm{q}'\pars{x}} & \ds{=} & \ds{\mrm{p}\pars{x}} \end{array}\right. \\[5mm] &\ \implies \mrm{p}'\pars{x} + \mrm{q}'\pars{x} = \mrm{p}\pars{x} + \mrm{q}\pars{x}\,,\quad \mrm{p}'\pars{x} - \mrm{q}'\pars{x} = -\bracks{\mrm{p}\pars{x} - \mrm{q}\pars{x}} \\[5mm] &\ \implies \mrm{p}\pars{x} + \mrm{q}\pars{x} = \expo{x}\,,\qquad\qquad\quad\,\,\,\,\, \mrm{p}\pars{x} - \mrm{q}\pars{x} = {1 \over \expo{x}} \\[5mm] & \implies \mrm{p}\pars{\ln\pars{2}} + \mrm{q}\pars{\ln\pars{2}}= 2\,,\qquad\quad\,\, \mrm{p}\pars{\ln\pars{2}} - \mrm{q}\pars{\ln\pars{2}}= {1 \over 2} \\[5mm] & \implies \mrm{p}\pars{\ln\pars{2}} = {5 \over 4}\,,\quad \mrm{q}\pars{\ln\pars{2}} = {3 \over 4} \implies \bbx{2^{\textsf{A}} = {5 \over 4}\,\textsf{1} + {3 \over 4}\textsf{A}} \end{align}

---

$$ \bbx{2^{\textsf{A}} = \pars{\begin{array}{rrr} \ds{1 \over 2} & \ds{-\,{3 \over 2}} & \ds{-\,{3 \over 2}} \\[1mm] \ds{3 \over 4} & \ds{11 \over 4} & \ds{3 \over 4} \\[1mm] \ds{-\,{3 \over 4}} & \ds{-\,{3 \over 4}} & \ds{5 \over 4} \end{array}}} $$

Answer 4

The characteristic polynomial of $A$ is $\lambda^3-\lambda^2-\lambda+1 = (\lambda-1)^2(\lambda+1)$, so by the Cayley-Hamilton theorem, every polynomial in $A$ can be reduced to a quadratic in $A$. This also holds for analytic functions of $A$ (in the region where everything converges, of course). In particular, $2^A$ can be expressed as a quadratic polynomial in $A$, i.e., $$2^A = a_0I+a_1A+a_2A^2 \tag{*}$$ for some as yet unknown coefficients $a_0$, $a_1$, $a_2$.

Now, if $f$ is analytic and $\lambda$ an eigenvalue of $A$, then $f(\lambda)$ is an eigenvalue of $f(A)$. The matrix in this problem has eigenvalues $1$ and $-1$, which gives us two linear equations for the unknown coefficients: $$a_0+a_1+a_2 = 2 \\ a_0-a_1+a_2=\frac12.$$ For a unique solution, we need one more independent equation, which we can generate by differentiation: $$a_1+2a_2=2\log2.$$ Solving this system and substituting back into (*) produces $$2^A = \left(\frac{13}8-\log2\right)I+\frac34 A+\left(\log2-\frac38\right)A^2 = \frac54I+\frac34A.$$

Answer 5

General method: to get $f(A)$ for a function $f$, find a polynomial that interpolates $f$ on the set of roots of a polynomial that annihilates $A$. (If the set has multiplicities, $P$ has also has to match the derivatives of $f$ at the points).

In our case, the polynomial $x^2-1$ annihilates $A$ and the roots of $x^2-1$ are $\pm 1$. So we need $P$ with $$P(1)=2^1 = 2\\ P(-1)=2^{-1}=1/2$$ The unique polynomial of degree $\le 1$ is $P(x)= 3/4 x + 5/4$. Therefore $$2^A= 3/4\cdot A + 5/4 I$$