Can a Linear Transformation that maps from R^m to R^n be Dot Product preserving?

Question

For $m \times n$ matrix $A$, textbook tries to prove (in the context of linear transformations) that following statements are equivalent:

(a) $A'A = I$

(b) $||Ax||=||x||$

(c) dot product of $Ax$ and $Ay$ = dot product of $x$ and $y$ for all $x$ and $y$ in $\mathbb{R}^n$

(d) column vectors of $A$ are orthonormal

Proof given by a textbook: click

I didnot understand part (c) => (d). They just define $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$... and prove it for that kind of transformation. Also, in (b)=>(c) part they just use theorem in which equivalence of (b) and (c) is proven for $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$. What I cannot understand is - why they assumed $m$ to be equal to $n$ for those parts?

Thank you

Answer 1

For $(c) \Rightarrow (d)$, take the canonic base of $\Bbb {R^n}, \{e_i\}, i=1,...,n$, now, we have that $=, \forall i,j=1,\ldots,n$, since the i-th column of the matrix is where the i-th base ends up. Now, by (c) and the previous statement: $$===0, \forall i,j=1,\ldots,n$$ Because the canonic base is orthogonal. So we conclude that the columns of A are orthogonal. To prove that they are orthonormal, we have that $$||A_{•i}||^2=||Ae_i||^2===||e_i||^2=1, \forall i=1,\ldots,n$$ Thus, the set $\{A_{•i}\}_{i=1}^n$ is orthonormal.