Orthogonal maps and orthonormal bases

Orthogonal and symmetric maps: Orthogonal maps

Orthogonal maps and orthonormal bases

According to Characterization of orthogonality by inner product, mutual inner products between two vectors do not change after transition to their images under orthogonal maps. In the following theorem we characterize these maps using orthonormal systems.

Let $V$ be a real inner product space. For a linear map $L:V\rightarrow V$ the following two statements are equivalent:

$L$ is orthogonal.
For each orthonormal system $\vec{a}_1,\ldots ,\vec{a}_n$ in $V$ , the system $L(\vec{a}_1),\ldots , L(\vec{a}_n)$ is also orthonormal.

$1.\Rightarrow 2.$ If $L$ is orthogonal, then, according to the Characterization of orthogonality by inner product, the inner product is invariant under $L$ . Let $\vec{a}_1,\ldots ,\vec{a}_n$ be an orthonormal system. Then, the inner product $\dotprod{L(\vec{a}_i)}{L(\vec{a}_j)}$ is equal to $\dotprod{\vec{a}_i}{\vec{a}_j}$ , and thus equal to $1$ if $i=j$ , and equal to $0$ if $i\neq j$ . This means that $L(\vec{a}_1),\ldots,L(\vec{a}_n)$ is an orthonormal system.

$2.\Rightarrow 1.$ Let $\vec{x}\in V$ . We will show that $\norm{L(\vec{x})}=\norm{\vec{x}}$ . If $\vec{x}=\vec{0}$ , this is true because the two sides are equal to $0$ . Suppose now that $\vec{x}$ is unequal to the zero vector. Then $\frac{1}{\norm{\vec{x}}}\vec{x}$ has length $1$ . So, this is an orthonormal system of one vector, that is, $\norm{L\left(\frac{1}{\norm{\vec{x}}}\vec{x}\right)}=1$ . We deduce from this: $\begin{array}{rcl}\norm{L(\vec{x})}&=&\norm{\norm{\vec{x}}\cdot L\left(\frac{1}{\norm{\vec{x}}}\vec{x}\right)}\\ &&\phantom{xx}\color{blue}{\text{linearity of }L}\\ &=&\norm{\vec{x}}\cdot \norm{L\left(\frac{1}{\norm{\vec{x}}}\vec{x}\right)}\\ &&\phantom{xx}\color{blue}{\text{multiplicativity of the norm}}\\ &=&\norm{\vec{x}}\cdot 1\\ &&\phantom{xx}\color{blue}{\text{equality above}}\\&=&\norm{\vec{x}}\end{array}$

The second statement says that an orthogonal map $L$ transforms orthonormal systems into orthonormal systems.

If $V$ is finite-dimensional, then we only need to study the images of one orthonormal system in order to conclude that a linear map is orthogonal.

Let $V$ be a real inner product space and suppose $V$ has finite dimension $m$ . For a linear map $L:V\rightarrow V$ the following three statements are equivalent:

$L$ is orthogonal.
For each orthonormal system $\basis{\vec{a}_1,\ldots ,\vec{a}_n}$ in $V$ the system $\basis{L(\vec{a}_1),\ldots , L(\vec{a}_n)}$ is orthonormal.
There is an orthonormal basis $\basis{\vec{a}_1,\ldots ,\vec{a}_m}$ of $V$ for which $\basis{L(\vec{a}_1),\ldots , L(\vec{a}_m)}$ is an orthonormal basis.

$1.\Rightarrow 2.$ This follows from the previous theorem.

$2.\Rightarrow 3.$ Choose an orthonormal basis $\basis{\vec{a}_1,\ldots ,\vec{a}_m}$ of $V$ (such a basis exists as a consequence of the Gram-Schmidt procedure). According to statement 2, the system $\basis{L(\vec{a}_1),\ldots ,L(\vec{a}_m)}$ is also orthonormal, and so (because of the first statement in theorem Properties of orthonormal systems of vectors) are independent. Because $m=\dim {V}$ , this system is a basis of $V$ .

$3.\Rightarrow 1.$ Let $\basis{\vec{a}_1,\ldots ,\vec{a}_m}$ be an orthonormal basis as in statement 3, so $\basis{L(\vec{a}_1),\ldots ,L(\vec{a}_m)}$ is an orthonormal basis. Furthermore, let $\vec{x}\in V$ . If $\vec{x}=\lambda_1 \vec{a}_1 +\cdots + \lambda_m \vec{a}_m$ , then $L(\vec{x})=\lambda_1 L(\vec{a}_1) + \cdots + \lambda_m L(\vec{a}_m)$ . Now we apply the Pythagorean theorem to both expressions (the square of the length of both $\lambda_i \vec{a}_i$ and $\lambda_i L(\vec{a}_i)$ is $\lambda_i^2$ )
$\norm{\vec{x}}^2=\lambda_1^2 + \cdots + \lambda_m^2 \quad \hbox{and} \quad \norm{L(\vec{x})}^2 = \lambda_1^2 + \cdots + \lambda_m^2$ which immediately implies $\norm{L(\vec{x})}=\norm{\vec{x}}$ .

An important consequence of this statement is that, if $\vec{x}$ and $\vec{y}$ are vectors of the same length in a finite-dimensional inner product space $V$ , then there is an orthogonal map $L:V\to V$ with $L(\vec{x}) = \vec{y}$ .

To see this, we assume that $\vec{x}$ and $\vec{y}$ are not equal to the zero vector (which we may for otherwise we can take $L$ to be the identity on $V$ ), extend $\frac{1}{\norm{\vec{x}}}\vec{x}$ to an orthonormal basis $\alpha$ of $V$ and $\frac{1}{\norm{\vec{y}}}\vec{y}$ to an orthonormal basis $\beta$ , and choose $L$ to be the linear map sending $\alpha$ to $\beta$ . Because of the implication $3\Rightarrow 1$ , the map $L$ is orthogonal.

Use It will become clear that the third statement is a strong criterion. When discussing matrices of orthogonal maps we will translate this statement into a property of the matrix of the linear map with respect to an orthonormal basis that is easy to verify.

We view $\mathbb{R}^3$ as the inner product space with standard inner product.

Does there exist a real number $a$ such that $L (\rv{1,0,0})=\frac{1}{21}\,\rv{a, -5, 20 }$ for an orthogonal map $L :\mathbb{R}^3\to \mathbb{R}^3$ ?

Yes

The map $L$ to be studied is linear, and so is orthogonal if and only if $\norm{L(\vec{x})}=\norm{\vec{x}}$ for all vectors $\vec{x}$ . In particular, we must have $\norm{L(\rv{1,0,0})}=\norm{\rv{1,0,0}}$ . This leads to the following equation with unknown $a$ , which we subsequently rewrite:
$\begin{array}{rcl}\norm{\dfrac{1}{21}\,\rv{a, -5, 20}}&=&1\\ &&\phantom{xx}\color{blue}{\text{mapping rule for }L\text{ used}}\\ \dfrac{1}{21}\cdot\norm{\rv{a, -5, 20}}&=&1\\ &&\phantom{xx}\color{blue}{\text{multiplicativity of the norm}}\\ \norm{\rv{a, -5, 20}}&=&21\\ &&\phantom{xx}\color{blue}{\text{both sides multiplied by }21}\\ a^2+(-5)^2+{20}^2&=&{21}^2\\ &&\phantom{xx}\color{blue}{\text{length computed and both sides squared}}\\ a^2&=&16\\ &&\phantom{xx}\color{blue}{\text{all constant terms carried to the right}}\\ a&=&\pm 4 \\ &&\phantom{xx}\color{blue}{\text{quadratic equation solved}} \end{array}$ Since $\rv{1,0,0}$ and $\frac{1}{21}\cdot \rv{4,-5,20}$ have equal length, the comment Extension of theorem Orthogonal maps on finite-dimensional inner product spaces and orthonormal systems shows that $L$ can be extended to an orthogonal map. So the answer is Yes.

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