### 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.

Orthogonal maps and 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.

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.

Orthogonal maps on finite-dimensional inner product spaces and orthonormal systems

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.

Does there exist a real number #a# such that \(L (\rv{1,0,0})=\frac{1}{31}\,\rv{a, 6, 30 }\) for an orthogonal map #L :\mathbb{R}^3\to \mathbb{R}^3#?

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}{31}\,\rv{a, 6, 30}}&=&1\\ &&\phantom{xx}\color{blue}{\text{mapping rule for }L\text{ used}}\\ \dfrac{1}{31}\cdot\norm{\rv{a, 6, 30}}&=&1\\ &&\phantom{xx}\color{blue}{\text{multiplicativity of the norm}}\\

\norm{\rv{a, 6, 30}}&=&31\\ &&\phantom{xx}\color{blue}{\text{both sides multiplied by }31}\\

a^2+{6}^2+{30}^2&=&{31}^2\\ &&\phantom{xx}\color{blue}{\text{length computed and both sides squared}}\\

a^2&=&25\\ &&\phantom{xx}\color{blue}{\text{all constant terms carried to the right}}\\

a&=&\pm 5 \\ &&\phantom{xx}\color{blue}{\text{quadratic equation solved}}

\end{array}\]Since #\rv{1,0,0}# and #\frac{1}{31}\cdot \rv{5,6,30}# 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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