### Orthogonal and symmetric maps: Symmetric maps

### Connection with symmetric matrices

The name *symmetric* for a linear map #L:V\to V# on a finite dimensional vector space #V# has to do with the matrix representation of #L#. We will use the following theorem to make the connection with matrices.

Characterizations of symmetryFor each linear map #L:V\rightarrow V#on a finite-dimensional inner product space #V# the following statements are equivalent:

- #L:V\rightarrow V# is symmetric.
- For each orthonormal system #\vec{a}_1,\ldots ,\vec{a}_n# in #V# we have #\dotprod{( L\vec{a}_i)}{\vec{a}_j}=\dotprod{\vec{a}_i }{(L \vec{a}_j)}# for all #i,j#.
- There is an orthonormal basis #\vec{a}_1,\ldots ,\vec{a}_m# of #V# with the property that #\dotprod{(L\vec{a}_i)}{ \vec{a}_j}=\dotprod{\vec{a}_i }{(L\vec{a}_j)}# for all #i,j#.
- There is an orthonormal basis #\vec{a}_1,\ldots ,\vec{a}_m# of #V# with the property that #\dotprod{(L\vec{a}_i)}{ \vec{a}_j}=\dotprod{\vec{a}_i }{(L\vec{a}_j)}# for all #i,j# with #i\lt j#.

Here is the link between *symmetric maps* and *symmetric matrices*.

Symmetric maps and matrices

Let #V# be a finite-dimensional real inner product space and #\alpha# an orthonormal basis of #V#. The linear map #L:V\rightarrow V# is symmetric if and only if the matrix #L_\alpha# of #L# relative to #\alpha# is symmetric.

#a =# #{3}#

In view of statement 4 of the theorem

\[\begin{array}{rcl} \dotprod{(6 + 3 x )}{{x}} &=& \dotprod{{1}}{(a+x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (6)\cdot(\dotprod{1}{x})+3 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+1\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\

3 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\

\end{array}\] We conclude that #a = 3#.

In view of statement 4 of the theorem

*Characterizations of symmetry*, it suffices to verify that \(\dotprod{(L({1}))}{{x}}= \dotprod{{1}}{(L({x}))}\). We work out this equation as follows\[\begin{array}{rcl} \dotprod{(6 + 3 x )}{{x}} &=& \dotprod{{1}}{(a+x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (6)\cdot(\dotprod{1}{x})+3 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+1\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\

3 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\

\end{array}\] We conclude that #a = 3#.

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