Here's a neat proof of 1. using matrix methods:
Consider
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix})
and its transpose
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix})
. Then
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A A^T = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \\ \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I)
where
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?I)
is the identity matrix. This implies that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T = A^{-1})
(so
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
is an
orthogonal matrix), and so we must have that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I = A^{-1} A = A^T A = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \\ \end{pmatrix})
so by equating entries, we obtain the result.
(Long) Edit: /0 gave me a major hint by showing that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?ad - bc = \pm 1)
, which is a necessary (but not sufficient) condition for the matrix
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix})
to be orthogonal. The proof above though shows that the conditions of the question are necessary and sufficient for
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
to be orthogonal. That is,
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\begin{array}{lll} a^2 + b^2 = 1 & \hspace{5.7cm} a^2 + c^2 = 1 \\ c^2 + d^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is orthogonal} \Longleftrightarrow b^2 + d^2 = 1 \\ ab + cd = 0 & \hspace{5.7cm} ac + bd = 0 \\ \end{array} )
This basically follows from the fact that a
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?n \times n)
matrix with real entries
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
,
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T A = I \Longleftrightarrow \left \langle Au , Av \right \rangle = \left \langle u , v \right \rangle \text{ for all } u,v \in \mathbb{R}^n)
.
Here
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\left \langle u , v \right \rangle = u^T v)
denotes the standard
inner product of two (column) vectors
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?u = \begin{pmatrix}u_1 \\ \vdots \\ u_n \\ \end{pmatrix}, v = \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} \in \mathbb{R}^n)
. It is closely related to the dot product;
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\left \langle u , v \right \rangle = u^T v = \begin{pmatrix}u_1 & \cdots & u_n \\ \end{pmatrix} \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} = u_1 v_1 + \ldots + u_n v_n = (u_1, \ldots, u_n) \cdot (v_1, \ldots, v_n).)
The linearity of the inner product means that we only need to show that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T A = I \Longleftrightarrow \left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle \text{ for all } 1 \leq i,j \leq n,)
where
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\{e_i : 1 \leq i \leq n\})
is some
basis for
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\mathbb{R}^n)
. We can clearly choose
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?e_1 = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0 \\ \end{pmatrix}, \ldots, e_n = \begin{pmatrix}0 \\ \vdots \\ 0 \\ 1 \\ \end{pmatrix})
, so that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\left \langle e_i , e_j \right \rangle = \delta_{ij})
, where
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\delta_{ij})
is the
Kronecker delta. We can then prove the result:
If
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T A = I)
, then
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\left \langle Ae_i , Ae_j \right \rangle = (Ae_i)^T(Ae_j) = e_i^T A^T A e_j = e_i^T e_j = \left \langle e_i , e_j \right \rangle.)
Conversely, if
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle)
, then
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\delta_{ij} = e_i^T (A^T A) e_j,)
which is the entry of the
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?i)
-th row and
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?j)
-th column of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T A)
, which implies that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^T A = I)
.
So the original question can be restated as:
If
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\delta_{ij} = e_i^T (A^T A) e_j)
, show that
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\delta_{ij} = e_i^T (A A^T) e_j)
, and the result holds because both expressions are equivalent to
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
being an orthogonal matrix. The result can clearly also be generalised to the
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?n \times n)
dimensional case, though of course it can't be written down so neatly.
There's also a similar problem for complex matrices; instead of considering orthogonal matrices, we consider
unitary matrices. A matrix with complex entries is unitary if
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A A^{*} = A^{*} A = I)
; here
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A^{*})
denotes the
conjugate transpose of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
, obtained by taking the transpose of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?A)
and conjugating each entry.
Thus the complex version of question 1. is
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\begin{array}{lll} |a|^2 + |b|^2 = 1 & \hspace{5.2cm} |a|^2 + |c|^2 = 1 \\ |c|^2 + |d|^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is unitary} \Longleftrightarrow |b|^2 + |d|^2 = 1 \\ a\overline{b} + c\overline{d} = \overline{a}b + \overline{c}d = 0 & \hspace{5.2cm} a\overline{c} + b\overline{d} = \overline{a}c + \overline{b}d= 0 \\ \end{array} )
(Can you tell I'm procrastinating from studying algebraic topology?
![Tongue :P](https://www.atarnotes.com/forum/Smileys/default/tongue.gif)
)