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5.A Orthogonal matrices

First, we will introduce orthogonal matrices, which are those corresponding to linear transformations that preserve all angles and lengths; in that sense they define a rigid motion in space. For example, any rotation of ${\mathbb{R}}^{n}$ around the origin doesn’t stretch any vectors, or change the angles between two vectors. So rotation matrices are examples of orthogonal matrices.

Theorem 5.1.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ be a square matrix. Then the following conditions are equivalent.

i.

$AA^{T}=I_{n}$ ,
ii.

$A^{T}A=I_{n}$ ,
iii.

The rows of $A$ form an orthonormal basis of ${\mathbb{R}}^{n}$ ,
iv.

The columns of $A$ form an orthonormal basis of ${\mathbb{R}}^{n}$ ,
v.

The linear transformation defined by $\vec{x}\mapsto A\vec{x}$ preserves the inner product; in other words, $\langle A\vec{x},A\vec{y}\rangle=\langle\vec{x},\vec{y}\rangle$ , for any $\vec{x},\vec{y}\in{\mathbb{R}}^{n}$ .

An orthogonal matrix is one which obeys any of the conditions in Theorem 5.1. To check whether a given matrix is orthogonal, only one (and any one) of the above conditions needs to be checked, since they are all equivalent.

Exercise 5.2:

For $R_{\theta}:=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$ , check each condition of Theorem 5.1.

[End of Exercise]

Proof of Theorem 5.1.

To prove that several different statements are equivalent, there are many different proof strategies that are logically valid. For example, a strategy different from the one used below would be to prove (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv) $\Rightarrow$ (v) $\Rightarrow$ (i). Whichever strategy is used, if one of the statements is assumed to be true, then all of the other statements must follow from it.

(i) $\Leftrightarrow$ (ii): This equivalence follows from the well-known fact that $AB=I_{n}$ implies $BA=I_{n}$ for square matrices over a field.

(i) $\Leftrightarrow$ (iii): Let $[A]_{ij}=a_{ij}$ be the coefficients of the matrix $A$ ; then $[A^{T}]_{ij}=a_{ji}$ . Using the formula for matrix multiplication (see Exercise 1.25), we obtain an expression for the $(i,j)$ entry of the matrix product:

[AA^{T}]_{ij}=\sum_{r=1}^{n}[A]_{ir}[A^{T}]_{rj}=\sum_{r=1}^{n}a_{ir}a_{jr}.

Since the $i$ th row of $A$ is the vector $(a_{i1},\cdots,a_{in})$ , the above sum is exactly the inner product of the $i$ th and $j$ th row. Therefore the rows are all orthonormal (and hence a basis) if and only if $[AA^{T}]_{ij}=1$ when $i=j$ and 0 otherwise; this is the same as $AA^{T}=I_{n}$ .

(ii) $\Leftrightarrow$ (iv) The columns of $A$ are orthonormal if and only if the rows of $A^{T}$ are orthonormal, so we apply the same argument as above, except replace $A$ with $A^{T}$ .

(ii) $\Leftrightarrow$ (v): By the definition of the standard inner product, for any $\vec{x},\vec{y}\in{\mathbb{R}}^{n}$ :

\langle A\vec{x},A\vec{y}\rangle=(A\vec{x})^{T}(A\vec{y})=(\vec{x}^{T}A^{T})(A% \vec{y})=\vec{x}^{T}(A^{T}A)\vec{y}.

The second equality used that $(AB)^{T}=B^{T}A^{T}$ , which is an elementary property of the transpose, seen in MATH105; the third equality used associativity of matrix multiplication. Now it is clear that if $A^{T}A=I_{n}$ then (v) is true. For the reverse implication, we use Exercise 5.3 to see that $A^{T}A-I_{n}=\mathbf{0}$ , as required. ∎

Exercise 5.3:

For $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ assume that

\vec{x}^{T}A\vec{y}=0

for all $\vec{x},\vec{y}\in{\mathbb{R}}^{n}$ . Prove that $A$ is the zero matrix. [See also: Theorem 3.6]

Exercise 5.4:

In Theorem 5.1, prove directly that (v) implies (iv).

[Hint: The columns of $A$ are $A\vec{e_{i}}$ , where $\vec{e_{i}}\in{\mathbb{R}}^{n}$ is a standard basis vector.]

[End of Exercise]