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 =# #{-9}#
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
\[\begin{array}{rcl} \dotprod{(-4 -9 x )}{{x}} &=& \dotprod{{1}}{(a+3 x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (-4)\cdot(\dotprod{1}{x})-9 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+3\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\
-9 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\
\end{array}\] We conclude that #a = -9#.
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
\[\begin{array}{rcl} \dotprod{(-4 -9 x )}{{x}} &=& \dotprod{{1}}{(a+3 x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (-4)\cdot(\dotprod{1}{x})-9 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+3\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\
-9 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\
\end{array}\] We conclude that #a = -9#.
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