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Question states: "Consider a linear transoformation $L$ from $\Bbb R^m$ to $R^n$ that preserves length. Let $A$ be the matrix of $L$... What is $A^TA$? What about $AA^T$?"

Here is what I have found out so far: We first notice that $A$ need not be an orthogonal matrix since $m$ need not equal to $n$. Since $L$ preserves legnth and the only vector that has length $0$ is the zero vector, $ker(L) = \{ 0 \}$. This means that $A$ has a leading 1 in every column, which implies that $dim(im(L)) = rank(A) =$ number of columns of $A = m$. That $A$ has a leading 1 in every column also implies that $n \ge m$. That $A$ has a leading 1 in every column implies that columns of $A$ are linearly independent. Also, $\{ 0 \} = ker(A) = ker(A^TA)$, so $A^TA$ is invertible. Also, if $m$ is strictly less than $n$, then the rref of $A^T$ has at least one column free, meaning that $A^Tx = 0$ has infinitely many solutions, meaning that $AA^Tx=0$ has infinitely many solutions and hence not invertible.

This is what I have found out so far. I was wondering whether $A^TA$ has to be the identity matrix. I've tried to figure out an counterexample, but failed.

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    While $A$ doesn't need to be square, it must have orthonormal columns. – 2017-02-21
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    Thanks so much. And why is that? – 2017-02-21
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    Because the standard unit vectors of $\Bbb R^m$ have to be sent to unit length, pairwise orthogonal vectors. – 2017-02-21

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We have for any two vectors $v,w\in\mathrm R^m$: \begin{align} v\cdot w &= (A v)\cdot (A w) && \text{by assumption}\\ &= (A v)^T (A w) && \text{definition of dot product}\\ &= v^T A^T A w && \text{property of the transpose} \end{align} Now consider the standard basis.