- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
7. $\quad$ If we interchange the rows and columns of a rectangular matrix $A,$ the new matrix so obtained is called the transpose of $A$ and is denoted by $A^t.$ For example, if we have
$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, \quad \text{then} \quad A^t = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}.$$
Prove that transposes have the following properties:
(a) $\quad$ $(A^t)^t = A.$
(b) $\quad$ $(A + B)^t = A^t + B^t.$
(c) $\quad$ $(cA)^t = cA^t.$
(d) $\quad$ $(AB)^t = B^tA^t.$
(e) $\quad$ $(A^{-1})^t = (A^t)^{-1}$ if $A$ is nonsingular.
(a) $\quad$ $(A^t)^t = A.$
$\quad$ Proof. $\quad$ Let $A$ be the $m \times n$ matrix given by $A = (a_{ij})_{i, j = 1}^{m,n}.$ Then, the transpose of $A$ is the $n \times m$ matrix given by $A^t = (a^t_{ij})_{i, j = 1}^{n,m},$ where for each pair $(i, j),$ the element $a^t_{ij} = a_{ji}.$ But if we apply the transpose operation to $A^t,$ this gives us the $m \times n$ matrix given by $(A^t)^t = (a^{tt}_{ij})_{i, j = 1}^{m,n},$ where for each pair $(i, j),$ the element $a^{tt}_{ij} = a^t_{ji} = a_{ij}.$ Hence, $(A^t)^t = A. \quad \blacksquare$
(b) $\quad$ $(A + B)^t = A^t + B^t.$
$\quad$ Proof. $\quad$ Let $A$ and $B$ be $m \times n$ matrices given by $A = (a_{ij})_{i, j = 1}^{m,n}$ and $B = (b_{ij})_{i, j = 1}^{m,n},$ with transposes $A^t$ and $B^t$ as defined in (a). By definition, the sum of $A$ and $B$ is $A + B = (a_{ij} + b_{ij})_{i, j = 1}^{m,n}.$ Then, its transpose is the $n \times m$ matrix $(A + B)^t$ whose $ij^{th}$ element $a^t_{ij} + b^t_{ij}$ is the sum $a_{ji} + b_{ji}$ for each pair $(i, j).$ But by definition of matrix addition, this means that $(A + B)^t = A^t + B^t. \quad \blacksquare$
(c) $\quad$ $(cA)^t = cA^t.$
$\quad$ Proof. $\quad$ Let $a_{ij}$ be the $ij^{th}$ element of $A$ and $ca_{ij}$ the $ij^{th}$ element of $cA$ for each pair $(i, j).$ Then, the $ij^{th}$ element of $A^t$ is $a_{ji}$ and the $ij^{th}$ element of $(cA)^t$ is $ca_{ji}.$ But by definition of scalar multiplication for matrices, this means $(cA)^t = cA^t. \quad \blacksquare$
(d) $\quad$ $(AB)^t = B^tA^t.$
$\quad$ Proof. $\quad$ By definition, if $A$ is an $m \times p$ matrix and $B$ is a $p \times n$ matrix, the matrix product $AB$ is the $m \times n$ matrix $C = (c_{ij})$ whose $ij^{th}$ entry is given by
\begin{align*}
c_{ij} &= \sum_{k = 1}^p a_{ik}b_{kj}
\end{align*}
In other words, the $ij^{th}$ entry of $AB$ is the dot product of the $i^{th}$
row of $A$ and the $j^{th}$ column of $B.$ Then, the transpose of $AB$ is the $n \times m$ matrix $C^t$ whose $ij^{th}$ element is $c_{ji},$ or:
\begin{align*}
c^t_{ij} &= \sum_{k = 1}^p a_{jk}b_{ki}
\end{align*}
In other words, the $ij^{th}$ element of $C^t$ is the dot product of the $j^{th}$ row of $A$ and the $i^{th}$ column of $B.$ Taking the transpose of $A$ and $B,$ the element $c^t_{ij}$ then becomes the dot product of the $i^{th}$ row of $B^t$ and the $j^{th}$ column of $A^t.$ In other words, if $A^t = (a^t_{ij})$ and $B^t = (b^t_{ij}),$ $c^t_{ij}$ is given by
\begin{align*}
c^t_{ij} &= \sum_{k=1}^p b^t_{ik}a^t_{kj}
\end{align*}
Hence, $C^t = (AB)^t = B^tA^t. \quad \blacksquare$
(e) $\quad$ $(A^{-1})^t = (A^t)^{-1}$ if $A$ is nonsingular.
$\quad$ Proof. $\quad$ From (d), we know that $(AB)^t = B^tA^t,$ hence if $A$ is nonsingular,
\begin{align*}
(AA^{-1})^t &= (A^{-1})^tA^t = I^t = I
\end{align*}
But because $(A^{-1})^tA^t = I,$ this makes $(A^{-1})^t$ the inverse of $A^t.$ That is,
\begin{align*}
(A^{-1})^t &= (A^t)^{-1}. \quad \blacksquare
\end{align*}