- 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
9. $\quad$ For each of the following statements about $n \times n$ matrices, give a proof or else exhibit a counter example.
(a) $\quad$ If $A$ and $B$ are orthogonal, then $A + B$ is orthogonal.
(b) $\quad$ If $A$ and $B$ are orthogonal, then $AB$ is orthogonal.
(c) $\quad$ If $A$ and $AB$ are orthogonal, then $B$ is orthogonal.
(a) $\quad$ If $A$ and $B$ are orthogonal, then $A + B$ is orthogonal.
$\quad$ Counterexample. $\quad$ Let $A = I$ and $B = -I.$ We can easily verify that $A$ and $B$ are orthogonal, but $A + B = O$ is not orthogonal. $\quad \blacksquare$
(b) $\quad$ If $A$ and $B$ are orthogonal, then $AB$ is orthogonal.
$\quad$ Proof. $\quad$ We wish to show that $AB(AB)^t = I.$ If $A$ and $B$ are orthogonal, then $AA^t = I$ and $BB^t = I.$ From Exercise 7, we also know that $(AB)^t = B^tA^t.$ As such, we have
\begin{align*}
AB(AB)^t &= AB(B^tA^t)
\\
&= A(BB^t)A^t
\\
&= A(I)A^t
\\
&= AA^t
\\
&= I. \quad \blacksquare
\end{align*}
(c) $\quad$ If $A$ and $AB$ are orthogonal, then $B$ is orthogonal.
$\quad$ Proof. $\quad$ If $AB$ is orthogonal, then $AB(AB)^t = I.$ But this means that $AB$ is the inverse of $(AB)^t.$ Hence, $(AB)^tAB = I,$ which gives us
\begin{align*}
(AB)^tAB &= (B^tA^t)AB = B^t(A^tA)B = I.
\end{align*}
Then, if $A$ is orthogonal, $AA^t = I,$ and $A^tA = I,$ giving us
\begin{align*}
B^t(A^tA)B &= B^tB = I
\end{align*}
But if this is the case, then $B^t$ is the inverse of $B,$ which means $BB^t = I.$ Thus, $B$ is orthogonal. $\quad \blacksquare$