# Operations I

• Nov 7th 2006, 09:33 AM
kgpretty
Operations I
I have a question for which I need help:

Let M consist of all 2*2 matrices of the form
Code:

[a    b] [-b  a]
Show that the product of the two matrices in M is a matrix in M.

Here's my attempt:
Code:

[a    b][a    b]    = [a^2-b^2          ab+ab] [-b  a][-b  a]      [-ab-ab        -b^2+a^2]                     = [a^2-b^2            2ab]                       [-2ab          -b^2+a^2]
.: the product of two matrices in M is a matrix is M. Simple enough!

However, the question goes on to say:

Let * denote the binary operation of matrix multiplication on M. Show that I is the identity element of M under *, and find the inverse of any arbitrary matrix of M.

Find two distinct elements x E M which satisfy x * x * x * x = I.

• Nov 7th 2006, 09:49 AM
ThePerfectHacker
Quote:

Originally Posted by kgpretty

Let * denote the binary operation of matrix multiplication on M. Show that I is the identity element of M under *, and find the inverse of any arbitrary matrix of M.

Find two distinct elements x E M which satisfy x * x * x * x = I.

If you drop the zero matrix then the determinant is $\displaystyle a^2+b^2>0$ hence all matrices are invertible.
Which means you have a group under matrix multiplication.

What you need is $\displaystyle x\in M$ such as,
$\displaystyle x^4=1$
Hence you need an element in $\displaystyle M$ whose order divides 4, that is 1,2,4.
The first case is simple.
The only element of order 1 is the identity element which satisfies that equation. Now just find an element of order two (and it will be different from the indentity element because it has different order).
That means,
Solve,
$\displaystyle \left[ \begin{array}{cc}a&b\\-b&a\end{array} \right]\cdot \left[ \begin{array}{cc}a&b\\-b&a\end{array} \right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$
$\displaystyle a^2-b^2=1$
$\displaystyle ab+ba=0$
$\displaystyle -ab-ba=0$
$\displaystyle a^2-b^2=1$
One such possibility is $\displaystyle a=-1,b=0$
Hence,
$\displaystyle \left[ \begin{array}{cc}-1&0\\0&-1\end{array}\right]$
(Remember to check because these are necessary though not suffienct conditions).
• Nov 7th 2006, 10:57 AM
topsquark
Quote:

Originally Posted by kgpretty
I have a question for which I need help:

Let M consist of all 2*2 matrices of the form
Code:

[a    b] [-b  a]
Show that the product of the two matrices in M is a matrix in M.

Here's my attempt:
Code:

[a    b][a    b]    = [a^2-b^2          ab+ab] [-b  a][-b  a]      [-ab-ab        -b^2+a^2]                     = [a^2-b^2            2ab]                       [-2ab          -b^2+a^2]
.: the product of two matrices in M is a matrix is M. Simple enough!

I should note that in your first problem you are merely showing, given a matrix $\displaystyle A \in M$, that $\displaystyle A^2 \in M$. What you need to do is define two matrices $\displaystyle A,B \in M$ and show that $\displaystyle AB \in M$ where A and B are not necessarily the same.

The proof is nearly identical to what you have done above.

-Dan
• Nov 7th 2006, 11:35 AM
Soroban
Hello, kgpretty!

Quote:

Let $\displaystyle M$ consist of all 2x2 matrices of the form: .$\displaystyle \begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$

Show that the product of the two matrices in $\displaystyle M$ is a matrix in $\displaystyle M$

You squared a matrix . . . We should use any two matrices.

$\displaystyle \begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix} * \begin{bmatrix}c & d \\ \text{-}d & c\end{bmatrix}\;=\;\begin{bmatrix}ac-bd & ad+bc \\ \text{-}bc -ad & \text{-}bd + ac\end{bmatrix} \;=\;\begin{bmatrix}ac-bd & ad + bc \\ \text{-}(ad + bc) & ac-bd\end{bmatrix}$

This product is of the proper form for $\displaystyle M.$
. . The two elements on the major diagonal are equal.
. . The two elements on the minor diagonal are equal with opposite signs.

Quote:

Let * denote the binary operation of matrix multiplication on $\displaystyle M$.
Show that $\displaystyle I$ is the identity element of $\displaystyle M$ under *.

$\displaystyle A*I \;= \;\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}*\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \;=\;\begin{bmatrix}a\!\cdot1\!+b\!\cdot\!0 & a\!\cdot\!0+b\!\cdot\!1\\ \text{-}b\!\cdot\!1+a\!\cdot\!0 & \text{-}b\!\cdot\!0 + a\!\cdot\!1\end{bmatrix}\;=\;\begin{bmatrix}a & b \\\text{-}b & a\end{bmatrix} \;= \;A$

Quote:

Find the inverse of any arbitrary matrix of $\displaystyle M.$

Let $\displaystyle A \,=\,\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$, and use your favorite method.

I happen to know that the inverse of $\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is: .$\displaystyle \begin{bmatrix}\frac{d}{ad-bc} & \frac{-b}{ad-bc} \\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}$

Therefore: .$\displaystyle A^{-1} \;= \;\begin{bmatrix}\frac{a}{a^2+b^2} & \frac{-b}{a^2+b^2} \\ \frac{b}{a^2+b^2} & \frac{a}{a^2+b^2}\end{bmatrix}$

Quote:

Find two distinct elements $\displaystyle x \in M$ which satisfy $\displaystyle x * x * x * x \:= \:I$

Let $\displaystyle x \:=\:\begin{bmatrix}a & b \\ \text{-}b & a\end{bmatrix}$

Then: .$\displaystyle x*x\:=\:\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}$

And: .$\displaystyle (x*x)*(x*x)\:=\:\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}*\begin{bmatrix}a^2-b^2 & 2ab \\ \text{-}2ab & a^2-b^2\end{bmatrix}$

. . . . . . . . . . . . . . . $\displaystyle = \:\begin{bmatrix}(a^2-b^2)^2 - 4a^2b^2 & 4ab(a^2-b^2) \\ \text{-}4ab(a^2-b^2) & (a^2-b^2)^2 - 4a^2b^2\end{bmatrix}$

Since: .$\displaystyle \begin{bmatrix}(a^2-b^2)^2 - 4a^2b^2 & 4ab(a^2-b^2) \\ \text{-}4ab(a^2-b^2) & (a^2-b^2)^2 - 4a^2b^2\end{bmatrix} \;=\;\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

. . we have: .$\displaystyle \begin{array}{cc}(A)\\(B)\end{array} \begin{array}{cc}(a^2-b^2)^2 - 4a^2b^2\:=\:1 \\ 4ab(a^2-b^2)\:=\:0\end{array}$

Equation (B) gives us three possible solutions:. $\displaystyle \left\{\begin{array}{ccc}(1)\;a\;= & 0 \\ (2)\;b\;= & 0 \\ (3)\;a\;= & \pm b\end{array}\right\}$

(1) If $\displaystyle a = 0$, (A) becomes: .$\displaystyle b^4 = 1\quad\Rightarrow\quad b = \pm1$

(2) If $\displaystyle b = 0$, (A) becomes: .$\displaystyle a^4 = 1\quad\Rightarrow\quad a = \pm1$

(3) If $\displaystyle a = \pm b$, (A) becomes: .$\displaystyle -4b^4 = 1$ ... which has no real roots.

There are four solutions: .$\displaystyle x\;= \;\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\;\;\begin{bmatrix}\text{-}1 & 0 \\ 0 & \text{-}1\end{bmatrix}\;\;\begin{bmatrix}0 & 1 \\ \text{-}1 & 0\end{bmatrix}\;\;\begin{bmatrix}0 & \text{-}1 \\ 1 & 0\end{bmatrix}$