Question

let V be a vector space of 3x3 symmetric matric.show that if A,B in V, then <A,B>=tr(AB) in inner product space

Answer 1

Let

$A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} &a_{32} & a_{33} \end{bmatrix}$ and $B=\begin{bmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} &b_{32} & b_{33} \end{bmatrix}$

$\Rightarrow AB=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} &a_{32} & a_{33} \end{bmatrix}\begin{bmatrix} b_{11} & b_{12} & b_{13}\\ b_{21} & b_{22} & b_{23}\\ b_{31} &b_{32} & b_{33} \end{bmatrix}$

$(AB) =\begin{bmatrix} a_{11}b_{11} +a_{21}b_{12} +a_{31}b_{13} & a_{11}b_{12} + a_{12}b_{22}+a_{13}b_{32}& a_{11}b_{13} + a_{12}b_{23}+a_{13}b_{33}\\ a_{21}b_{11} + a_{22}b_{21}+a_{23}b_{31}& a_{21}b_{12} + a_{22}b_{22}+a_{23}b_{32} & a_{21}b_{13} + a_{22}b_{23}+a_{23}b_{33}\\ a_{31} b_{11} + a_{32}b_{21}+ a_{33}b_{31}&a_{31} b_{12} + a_{32}b_{22}+ a_{33}b_{32}&a_{31} b_{13} + a_{32}b_{23}+ a_{33}b_{33} \end{bmatrix}$

$(AB)^T =\begin{bmatrix} a_{11}b_{11} +a_{21}b_{12} +a_{31}b_{13} & a_{21}b_{11} + a_{22}b_{21}+a_{23}b_{31}& a_{31} b_{11} + a_{32}b_{21}+ a_{33}b_{31}\\ a_{11}b_{12} + a_{12}b_{22}+a_{13}b_{32}& a_{21}b_{12} + a_{22}b_{22}+a_{23}b_{32} &a_{31} b_{12} + a_{32}b_{22}+ a_{33}b_{32} \\ a_{11}b_{13} + a_{12}b_{23}+a_{13}b_{33}&a_{21}b_{13} + a_{22}b_{23}+a_{23}b_{33}&a_{31} b_{13} + a_{32}b_{23}+ a_{33}b_{33} \end{bmatrix}$

$<A ,B>=a_{11}b_{11} +a_{21}b_{12} +a_{31}b_{13} + a_{21}b_{11} + a_{22}b_{21}+a_{23}b_{31}+ a_{31} b_{11} + a_{32}b_{21}+ a_{33}b_{31} a_{11}b_{12} + a_{12}b_{22}+a_{13}b_{32}+ a_{21}b_{12} + a_{22}b_{22}+a_{23}b_{32} +a_{31} b_{12} + a_{32}b_{22}+ a_{33}b_{32} a_{11}b_{13} + a_{12}b_{23}+a_{13}b_{33}+a_{21}b_{13} + a_{22}b_{23}+a_{23}b_{33}+a_{31} b_{13} + a_{32}b_{23}+ a_{33}b_{33}$

$=\Sigma_{i=1}^{i=3}\Sigma_{j=1}^{j=3}a_{ij}b_{ij}$