Orthonormal vectors and orthogonal matrices

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An orthogonal set of vectors $\{{\bf u}_1, {\bf u}_2,\dots ,{\bf u}_n\}$ is said to be orthonormal if $\|{\bf u}_i\| = 1, 1\le i\le n$ . Clearly, given an orthogonal set of vectors $\{{\bf v}_1, {\bf v}_2,\dots ,{\bf v}_n\}$ , one can orthonormalize it by setting ${\bf u}_i = {\bf v}_i/\|{\bf v}_i\|$ for each $i$ . Orthonormal bases in $\mathbb R^n$ “look” like the standard basis, up to rotation of some type.

We call an $n\times n$ matrix $A$ orthogonal if the columns of $A$ form an orthonormal set of vectors¹ .

Show that an $n\times n$ matrix $A$ is orthogonal iff $A^T*A = I$ .

An $n\times n$ matrix $A$ is orthogonal iff ${\bf v}\cdot {\bf w} = (A*{\bf v})\cdot (A*{\bf w})$ for all ${\bf v}, {\bf w}\in \mathbb R^n$ .

Proof: By Theorem thm:matrep, we see that two matrices $A,B$ satsify the property ${\bf v}^T*A^T*A*{\bf w} = (A*{\bf v})\cdot (A*{\bf w}) = (B*{\bf v})\cdot (B*{\bf w}) = {\bf v}^T*B^T*B*{\bf w}\qquad \forall {\bf v}, {\bf w}\in \mathbb R^n$ iff $A^T*A=B^T*B$ . The hypothesis of the lemma can be restated as $(A*{\bf v})\cdot (A*{\bf w}) = (I*{\bf v})\cdot (I*{\bf w})\qquad \forall {\bf v}, {\bf w}\in \mathbb R^n$ implying $A^T*A=I^T*I = I$ , which by the previous exercise is equivalent to $A$ being orthogonal. $\blacksquare$

The notion of orthogonality for matrices is a special example of a linear transformation preserving a given inner product, which we will discuss in more detail below.