Answer to Question #146301 in Discrete Mathematics for Promise Omiponle

Question

Let A be a square matrix, and AT denotes the transpose of A. Show that the following hold true.
(a) (AT)T=A
(b) (A+B)T=AT+BT
(c) (AB)T=BT AT.

Answer 1

Let $A=\left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array} \right]$ and $B=\left[ \begin{array}{cccc} b_{11} & b_{12} & \ldots & b_{1n}\\ b_{21} & b_{22} & \ldots & b_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{array} \right]$.

(a) $A^T=\left[ \begin{array}{cccc} a_{11} & a_{21} & \ldots & a_{n1}\\ a_{12} & a_{22} & \ldots & a_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ a_{1n} & a_{2n} & \ldots & a_{nn} \end{array} \right]$ and $(A^T)^T=\left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array} \right]=A.$

(b) $(A+B)^T=\left(\left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array} \right]+ \left[ \begin{array}{cccc} b_{11} & b_{12} & \ldots & b_{1n}\\ b_{21} & b_{22} & \ldots & b_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{array} \right]\right)^T=$

$=\left[ \begin{array}{cccc} a_{11}+b_{11} & a_{12}+b_{12} & \ldots & a_{1n}+b_{1n}\\ a_{21}+b_{21} & a_{22}+b_{22} & \ldots & a_{2n}+b_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1}+b_{n1} & a_{n2}+b_{n2} & \ldots & a_{nn}+b_{nn} \end{array} \right]^T=$

$= \left[ \begin{array}{cccc} a_{11}+b_{11} & a_{21}+b_{21} & \ldots & a_{n1}+b_{n1}\\ a_{12}+b_{12} & a_{22}+b_{22} & \ldots & a_{n2}+b_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ a_{1n}+b_{1n} & a_{2n}+b_{2n} & \ldots & a_{nn}+b_{nn} \end{array} \right]=$

$=\left[ \begin{array}{cccc} a_{11} & a_{21} & \ldots & a_{n1}\\ a_{12} & a_{22} & \ldots & a_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ a_{1n} & a_{2n} & \ldots & a_{nn} \end{array} \right]+ \left[ \begin{array}{cccc} b_{11} & b_{21} & \ldots & b_{n1}\\ b_{12} & b_{22} & \ldots & b_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ b_{1n} & b_{2n} & \ldots & b_{nn} \end{array} \right]=A^T+B^T.$

(c) $(AB)^T=\left(\left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array} \right]\cdot \left[ \begin{array}{cccc} b_{11} & b_{12} & \ldots & b_{1n}\\ b_{21} & b_{22} & \ldots & b_{2n}\\ \ldots & \ldots & \ldots & \ldots\\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{array} \right]\right)^T=$

$=\left[ \begin{array}{cccc} a_{11}b_{11}+\cdots+a_{1n}b_{n1} & a_{11}b_{12}+\cdots+a_{1n}b_{n2} & \ldots & a_{11}b_{1n}+\cdots+a_{1n}b_{nn}\\ a_{21}b_{11}+\cdots+a_{2n}b_{n1} & a_{21}b_{12}+\cdots+a_{2n}b_{n2} & \ldots & a_{21}b_{1n}+\cdots+a_{2n}b_{nn}\\ \ldots & \ldots & \ldots & \ldots\\ a_{n1}b_{11}+\cdots+a_{nn}b_{n1} & a_{n1}b_{12}+\cdots+a_{nn}b_{n2} & \ldots & a_{n1}b_{1n}+\cdots+a_{nn}b_{nn}\\ \end{array} \right]^T=$

$=\left[ \begin{array}{cccc} a_{11}b_{11}+\cdots+a_{1n}b_{n1} & a_{21}b_{11}+\cdots+a_{2n}b_{n1} & \ldots & a_{n1}b_{11}+\cdots+a_{nn}b_{n1}\\ a_{11}b_{12}+\cdots+a_{1n}b_{n2} & a_{21}b_{12}+\cdots+a_{2n}b_{n2} & \ldots & a_{n1}b_{12}+\cdots+a_{nn}b_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ a_{11}b_{1n}+\cdots+a_{1n}b_{nn} & a_{21}b_{1n}+\cdots+a_{2n}b_{nn} & \ldots & a_{n1}b_{1n}+\cdots+a_{nn}b_{nn}\\ \end{array} \right]=$

$= \left[ \begin{array}{cccc} b_{11} & b_{21} & \ldots & b_{n1}\\ b_{12} & b_{22} & \ldots & b_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ b_{1n} & b_{2n} & \ldots & b_{nn} \end{array} \right] \cdot \left[ \begin{array}{cccc} a_{11} & a_{21} & \ldots & a_{n1}\\ a_{12} & a_{22} & \ldots & a_{n2}\\ \ldots & \ldots & \ldots & \ldots\\ a_{1n} & a_{2n} & \ldots & a_{nn} \end{array} \right] =B^T A^T.$