Prove / Provide counter example - about inner product, orthogonal matrix

Question

Let $A ∈ M_n (\Bbb R)$ and let $(•, •)$ be the standard inner product on $\Bbb R^n$.

1. If for all $u \in \Bbb R^n$ we have $||Au||=||u||$ then $A$ is an orthogonal matrix.
2. If for all $\Bbb v,u \in \Bbb R^n$ such that $\Bbb v \perp u$ we have $A \Bbb v \perp Au$ then $A$ is an orthogonal matrix.

I have proved the other side for both of those statements (if $A$ is orthogonal, then...), but I'd like to know what happens from the other direction.

Thanks a lot!

Answer 1

Statement (1) relies on the following fact:

Fact: If $X\in\mathbb{R}^{n\times n}$ is symmetric, and if $u^TXu=0$ for all $u\in \mathbb{R}^n$, then $X=O$.

Proof: Write $X=QDQ^T$ where $D=\text{diag}(\lambda_1,\ldots,\lambda_n)$ and $Q$ is orthogonal. Fix any $j\in \{1,...,n\}$ and put $u:=Qe_j$. Then we have $$\lambda_j=e_j^TDe_j=e_j^TQ^TXQe_j=u^TXu=0$$ This shows $D=O$ and hence so is $X$.

Now if $\|Au\|=\|u\|$ for all $u$ then $u^T\Big(A^TA-I\Big)u=0$ for all $u$. Since $A^TA-I$ is symmetric, we must have $A^TA-I=O$ i.e. $A^TA=I$ and we're done.

Answer 2

The defining property of an orthogonal matrix is that it preserves inner products. In $\mathbb{R}^n$, the inner product is just $(u, v) = u^Tv$, viewing $u$ and $v$ as column vectors.

Hence, $(Au, Av) = u^TA^TAv$. In the case of $u = v$, we have the situation you described in part 1.

Preserving the norm means $A$ satisfies $u^TA^TAu = u^Tu$. This equation holds for every vector $u$, including the basis vectors. Choose an orthonormal basis, $\{e_i\}_{i = 1}^n$.

Hence, $e_i^TA^TAe_i = 1$ for every basis vector.

But since we have an orthonormal basis, this equation just says that the component of $A^TAe_i$ along the $e_i$ direction is preserved, i.e. equals 1. Since this holds for all $i$, the component along every basis vector is preserved by $A^TA$.

But this is nothing but saying that $A^TA$ is the identity, i.e. $A$ is orthogonal.

This proves the direction you were asking for in part 1.