List Two Different Matrices A s.t. A^2=I

**Fourier** · September 4th 2007, 04:35 PM

Hello,

I am trying to determine two different matrices $A$ such that $A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg]$ .

I am trying to solve this algebraically. Here is what I have attempted:

**JakeD** · September 4th 2007, 06:39 PM

Originally Posted by Fourier

Hello,

I am trying to determine two different matrices $A$ such that $A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg]$ .

I am trying to solve this algebraically. Here is what I have attempted:

Set b = c = 0. Then a^2 = d^2 = 1. So I and -I fall out as 2 of 4 solutions.

**CaptainBlack** · September 4th 2007, 07:19 PM

Originally Posted by Fourier

Hello,

I am trying to determine two different matrices $A$ such that $A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg]$ .

I am trying to solve this algebraically. Here is what I have attempted:

Set $a^2=d^2=0$ , then $bc=1$ so:

$<br /> A^2= \left[ \begin{array}{cc}0&x\\1/x&0 \end{array} \right]^2=I_{2x2} <br />$

RonL

**Soroban** · September 5th 2007, 05:10 AM

Hello, Fourier!

You're off to a good start . . .

. . $\begin{bmatrix}a & b\\c &d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix }\;=\;\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\begin{array}{ccc}(1)\;\;a^2 + bc \:=\:1 &\qquad & (2)\;\;ab+bd \:=\:0 \\<br /> <br /> (3)\;\;ac + cd \:=\:0 & \qquad & (4)\;\;bc+d^2\:=\:0\end{array}$

We have: . $\begin{array}{cccc}(2)\;\;ab+bd\:=\:0 & \Rightarrow & b(a+d)\:=\:0 & (5) \\ (3)\;\;ac + cd \:=\:0 & \Rightarrow & c(a+d)\:=\:0\ & (6)\end{array}$

Subtract: . $\begin{array}{c}(1)\;\;a^2+bc \:=\:1 \\ (4)\;\;bc+d^2\:=\:1\end{array} \quad\Rightarrow\quad a^2-d^2\:=\:0\quad\Rightarrow\quad d \,=\,\pm a$

If $a\!\cdot\!d \neq 0$ , then (5) and (6) give us: . $b = c = 0$

And (1) and (4) give us: . $a^2 \,= \,1,\;d^2\,=\,1\quad\Rightarrow\quad a\,=\,\pm1,\;d\,=\,\pm1$

. . Two solutions: . $\begin{bmatrix}1 &0 \\0&1\end{bmatrix}$ .and . $\begin{bmatrix}\text{-}1 & 0\\0&\text{-}1\end{bmatrix}$

If $a = d = 0$ , then (1) gives us: . $bc \,=\,1\quad\Rightarrow\quad c \,=\,\frac{1}{b}$

. . More solutions: . $\begin{bmatrix}0 & b \\ \frac{1}{b} & 0\end{bmatrix}$ . . . . for $b \neq 0$ . obviously.

**JakeD** · September 5th 2007, 06:02 AM

Originally Posted by Soroban

Hello, Fourier!

You're off to a good start . . .

We have: . $\begin{array}{cccc}(2)\;\;ab+bd\:=\:0 & \Rightarrow & b(a+d)\:=\:0 & (5) \\ (3)\;\;ac + cd \:=\:0 & \Rightarrow & c(a+d)\:=\:0\ & (6)\end{array}$

Subtract: . $\begin{array}{c}(1)\;\;a^2+bc \:=\:1 \\ (4)\;\;bc+d^2\:=\:1\end{array} \quad\Rightarrow\quad a^2-d^2\:=\:0\quad\Rightarrow\quad d \,=\,\pm a$

If $a\!\cdot\!d \neq 0$ , then (5) and (6) give us: . $b = c = 0$

And (1) and (4) give us: . $a^2 \,= \,1,\;d^2\,=\,1\quad\Rightarrow\quad a\,=\,\pm1,\;d\,=\,\pm1$

. . Two solutions: . $\begin{bmatrix}1 &0 \\0&1\end{bmatrix}$ .and . $\begin{bmatrix}\text{-}1 & 0\\0&\text{-}1\end{bmatrix}$

If $a = d = 0$ , then (1) gives us: . $bc \,=\,1\quad\Rightarrow\quad c \,=\,\frac{1}{b}$

. . More solutions: . $\begin{bmatrix}0 & b \\ \frac{1}{b} & 0\end{bmatrix}$ . . . . for $b \neq 0$ . obviously.

More solutions are

$\begin{bmatrix}\text{-1} &0 \\0&1\end{bmatrix}$ and $\begin{bmatrix}1 & 0\\0&\text{-}1\end{bmatrix}$

and

$\begin{bmatrix}a & b \\ \frac{1-a^2}{b} & \text{-}a\end{bmatrix}$ and $\begin{bmatrix}a & \frac{1-a^2}{b} \\ b & \text{-}a\end{bmatrix}$ for $b \neq 0.$

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