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

• Sep 4th 2007, 04:35 PM
Fourier
List Two Different Matrices A s.t. A^2=I
Hello,

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

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

http://img207.imageshack.us/img207/7...4194511xi6.jpg
• Sep 4th 2007, 06:39 PM
JakeD
Quote:

Originally Posted by Fourier
Hello,

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

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

http://img207.imageshack.us/img207/7...4194511xi6.jpg

Set b = c = 0. Then a^2 = d^2 = 1. So I and -I fall out as 2 of 4 solutions.
• Sep 4th 2007, 07:19 PM
CaptainBlack
Quote:

Originally Posted by Fourier
Hello,

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

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

http://img207.imageshack.us/img207/7...4194511xi6.jpg

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

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

RonL
• Sep 5th 2007, 05:10 AM
Soroban
Hello, Fourier!

You're off to a good start . . .

Quote:

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

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

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

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

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

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

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

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

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

• Sep 5th 2007, 06:02 AM
JakeD
Quote:

Originally Posted by Soroban
Hello, Fourier!

You're off to a good start . . .

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

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

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

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

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

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

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

More solutions are

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

and

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