The assignment:
a) Prove that square-matrix A is orthogonal if and only if A has orthonormal columns.
b) Prove that square-matrix A is orthogonal if and only if A has orthonormal rows.
So I know that A matrix has orthonormal columns if and if only $A^TA=I$.
But how about orthonormal rows? Should I use $AA^T=I$ ?
b) For example, can I prove like this (?) :
Let be $A=\begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}$
$AA^T=I$
$AA^T=\begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} \times \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}^T = \begin{bmatrix} a_1a_1^T & a_2a_2^T & a_3a_3^T \end{bmatrix}$
$a_1a_1^T=1 \quad\quad a_2a_2^T=1 \quad\quad a_3a_3^T=1$
So $A$ is orthogonal, because rows of matrix A are orthonormal. $\Box$
a) I did it like this which I think is correct:
$A^TA=I$
$A^TA=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}^T \times \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} =
\begin{bmatrix}
a_1^Ta_1 & a_1^Ta_2 & a_1^Ta_3
\\
a_2^Ta_1 & a_2^Ta_2 & a_2^Ta_3
\\
a_3^Ta_1 & a_3^Ta_2 & a_3^Ta_3
\end{bmatrix}$
$a_1^Ta_2=0 \quad a_1^Ta_3=0$
$a_2^Ta_1=0 \quad a_2^Ta_3=0$
$a_3^Ta_1=0 \quad a_3^Ta_2=0$
$a_1^Ta_1=1 \quad\quad a_2^Ta_2=1 \quad\quad a_3^Ta_3=1$
So $A$ is orthogonal, because columns of matrix A are orthonormal. $\Box$
Best Answer
For (b): Let me denote the matrix $A$ as follows: $$\begin{pmatrix} - & a_1 & -\\ - & a_2 & -\\ & \vdots & \\ - & a_n & - \end{pmatrix}$$ where the $a_i$ are row vectors and I emphasised this by adding '-'. We know that a matrix $A$ is orthogonal if $AA^T = I$. We want to show that the rows of $A$ form an orthonormal set, so let us take two arbitrary rows, $a_j$ and $a_k$, with $1 \leq j,k \leq n$. Note that we have that $$AA^T = \begin{pmatrix} - & a_1 & -\\ - & a_2 & -\\ & \vdots & \\ - & a_n & - \end{pmatrix}\begin{pmatrix} | & | & & | \\ a_1^T & a_2^T & \ldots & a_n^T\\ | & | & & | \end{pmatrix} = I$$ so if we compute $a_ja_k^T$, this corresponds to the entry in row $j$, column $k$ of the identity matrix. This entry is equal to $0$ if $j \neq k$ and equal to $1$ if $j = k$. This shows that the rows of $A$ form an orthonormal set. The other implication (orhtonormal rows implies $A$ orthogonal) follows in the same way.