A matrix $A \in \operatorname{Mat}(n \times n, \Bbb R)$ is said to be orthogonal if its columns are orthonormal relative to the dot product on $\Bbb R^n$.
- By considering $A^TA$, show that $A$ is an orthogonal matrix if and only if $A^T = A^{−1}$.
- Deduce that the rows of any $n × n$ orthogonal matrix $A$ form an orthonormal basis for the space of $n$-component row vectors over $\Bbb R$.
I am trying to do part 2.
What I tried is that since we figured out that $A^T = A^{-1}$, and the inverse of $A$ is the left product of elementary matrices to $A$, the row space of $A^TA =$ row space of $A$.
Also, since $A^TA = I$, a basis of the row space of $A$ is a basis of the row space of $I$. Since the columns of $I$ are the standard basis of $\Bbb R^n$ $(e_1, …, e_n)$, and are orthonormal to each other, they form an orthonormal basis of $\Bbb R^n$.
Something tells me this proof is wrong. Could someone give me some guidance?
Best Answer
Say the columns of $A$ are $A_1,\dots,A_n$.
If $B = (b_{ij}) = A^TA$, then:
$b_{ij} = \langle (A_i)^T,A_j\rangle$, and since these (the $A_k$) are orthonormal relative to our inner product, we have:
$b_{ij} = \delta_{ij}$ (the kronecker delta), which is $1$ when $i = j$, and $0$ otherwise, that is to say $B$ is the identity matrix.
So $A^TA = I$, from which we conclude $A^T = A^{-1}$ (If you insist on showing $A$ has a two-sided inverse, see below).
On the other hand, if $A^TA = I$, then (running our argument in reverse), we see the columns of $A$ (and thus the rows of $A^T$) form an orthonormal basis (they form a basis since $A$ is invertible).
By considering $AA^T = I$ in the same way, we see the columns of $A^T$, and thus the rows of $(A^T)^T = A$ also form an orthonormal basis.