91Ó°ÊÓ

The two given matrices are A and B.

Consider the three matrices A, B and C as,

$A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]andC=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

The matrix S is a symmetric matrix and A is an antisymmetric matrix, which means$S=S^T\text{and}A=-A^T.$

A symmetric matrix is a square matrix that is identical to its transpose in linear algebra.

An antisymmetric matrix is a square matrix whose negative transpose equals its positive transpose.

a)
By using the definition of trace of a matrix in index notation,

$$\begin{gathered} Tr\left(AB\right)=\underset{i}{∑}{\left(AB\right)}_{ii} \\ =\underset{ij}{∑}\left(a_{ij}{b}_{ji}\right) \\ =\underset{ij}{∑}\left(b_{ji}{a}_{ij}\right) \\ =\underset{j}{∑}{\left(BA\right)}_{jj} \\ =Tr\left(BA\right) \end{gathered}$$

Hence, it is proved that$Tr\left(AB\right)=Tr\left(BA\right)$.

b)

Consider the three matrices A, B, and C as shown below.

$A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],C=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$

Taking product of ABC, ACB and CAB respectively,

localid="1664255251275" $$\begin{gathered} ABC=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right] \\ CAB=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right] \\ =\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \\ ACB=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right] \\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

Now, $Tr\left(ABC\right)=1,Tr\left(ACB\right)=0$, and$Tr\left(CAB\right)=1$

Hence, proved.

c)

Matrixis a symmetric matrix andis an antisymmetric matrix, which means that

$S=S^T\text{and}A=-A^T.$

Evaluating$Tr(SA{)}^T$,

$$\begin{gathered} Tr(SA{)}^T=Tr\left(A^T{S}^T\right)....\left(U\sin gpropertyofsymmetricandantisymmetricmatrices\right) \\ =Tr\left(\right(-A\left)\right(S\left)\right) \\ =-Tr\left(SA\right)……\left(1\right) \end{gathered}$$

Since, the trace of a transpose matrix is same as that of the original matrix.

Therefore, plugging for$Tr(SA{)}^T=Tr\left(SA\right)$in (1)

$$\begin{gathered} Tr\left(SA\right)=-Tr\left(SA\right) \\ 2Tr\left(SA\right)=0 \\ Tr\left(SA\right)=0 \end{gathered}$$

Hence, proved.