91Ó°ÊÓ

Demonstration of the given statement.

In abstract algebra, the symmetric group defined over any set is the group whose elements are all the bisections from the set to itself, and whose group operation is the composition of functions.

Since, in question a given a matrix has been denoted by C so denote diagonalize matrix by P, and show that $P^{-1}AP$is a diagonal matrix. Thus, matrix A can be diagonalized by finding an invertible matrix P.

Here, first find matrix P for matrix $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$by find its eigenvalues and eigenvectors as:

Step 1: Put $$\begin{gathered} \left(A-\mathrm{λ}D\right)=0 \\ \left|\begin{array}{cc}0-\mathrm{λ}& 1\\ 1& 0-\mathrm{λ}\end{array}\right|=0 \\ {\mathrm{λ}}^2-1=0⇒\mathrm{λ}=1 \end{gathered}$$, - .

Step2: The eigenvectors corresponding to the eigenvalue role="math" localid="1659083743038" $\mathrm{λ}=1$are the solutions of equation role="math" localid="1659083787913" $\left(A-1·l\right)X=0$where $X={\left(x_1.x_2\right)}^{-1}$

$$\begin{gathered} \left(\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)-x_1+x_2=0,x_1-x_2=0 \\ {x}_1=-x_2=k( \end{gathered}$$A real number $)$.

Thus, $X=k\left(\begin{array}{c}1\\ 1\end{array}\right)$.

Similarly, the eigenvectors corresponding to the eigenvalue $\mathrm{λ}=-1$are the solutions of equation where

$$\begin{gathered} \left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)x_1+x_2=0,x_1+x_2=0 \\ {x}_1=-x_2\stackrel{→}{AB} \end{gathered}$$, so if $x_1=k$then $x_2=-k$A real number

Thus, $X=\left(\begin{array}{c}1\\ -1\end{array}\right)$

The diagonalize is $P^{-1}=\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)$

$âˆ\mu$If $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$then $M^{-1}=\frac{1}{∆d-b}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$.

Now, use matrix P we diagonalize each of 4 given matrix as:

Diagonalization of matrix I:

$$\begin{gathered} P^{-1}P=\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)·\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)·\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)·\left(\begin{array}{c}1·1+\\ 0·1+1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)·\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1·1+-1·1& -1·1+-1·-1\\ -1·1+-1·1& -1·1+1·-1\end{array}\right) \end{gathered}$$

Then

$\begin{array}{l}=\frac{1}{2}\left(-\begin{array}{cc}2& 0\\ 0& -2\end{array}\right)\\ =-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}$

That, is diagonalization of matrix I that was already an identity matrix.

Diagonalization of matrix A :

$$\begin{gathered} P^{-1}AP=\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)·\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)·\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)·\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right) \end{gathered}$$

$$\begin{gathered} \left(\begin{array}{cc}0·1+1·1& 0·1+1·-1\\ 1·1+0·1& 1·1+0·-1\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}-1·1+-1·1& -1·-1+1·-1\\ -1·1+1·1& -1·-1+0·-1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

After find the diagonalization

Since, the matrices I, A, B, and are the elements of a group so the matrix P obtained for each elements will diagonalize the other remaining matrices.

The matrices I, A, B, and C that are reducible representation of a group can be diagonalized by finding a diagonalizing matrix.

Hence, to demonstrate.