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.
For 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#.
The implications #1\Rightarrow 2#, #2 \Rightarrow 3#, and #3 \Rightarrow 4# are trivial.
#3\Rightarrow 1#: Write #\vec{x}=\lambda_ 1\vec{a}_1 + \cdots + \lambda_m\vec{a}_m# and #\vec{y}=\mu_ 1\vec{a}_1 + \cdots + \mu_m\vec{a}_m#. Then
\[
\dotprod{(A \vec{x})}{\vec{y}}=\sum_{i=1}^m \sum_{j=1}^m\lambda_i \mu_j \dotprod{(A\vec{a}_i) }{\vec{a}_j}=
\sum_{i=1}^m \sum_{j=1}^m\lambda_i \mu_j \dotprod{\vec{a}_i }{(A\vec{a}_j)}=
\dotprod{ \vec{x}}{(A \vec{y})}
\]
#4\Rightarrow 3#: The proof of this implication follows from the following two observations:
- If #i=j# then #\dotprod{(L\vec{a}_i)}{ \vec{a}_j}=\dotprod{\vec{a}_i }{(L\vec{a}_j)}# is a direct consequence of the symmetry of the inner product.
- If #i\gt j# then #\dotprod{(L\vec{a}_i)}{ \vec{a}_j}=\dotprod{\vec{a}_i }{(L\vec{a}_j)}# is a consequence of #\dotprod{(L\vec{a}_j)}{ \vec{a}_i}=\dotprod{\vec{a}_j }{(L\vec{a}_i)}#.
Each linear map #\mathbb{R}\to \mathbb{R}# is a scalar multiplication. This is always symmetric. If the scalar that is multiplied by equals #\lambda#, then the corresponding matrix is #\matrix{\lambda}#; each #(1\times1)#-matrix is indeed always symmetric.
Suppose that #\dim{V} = 2 # and that #\basis{\vec{a}_1,\vec{a}_2}# is an orthonormal basis of #V#. Then, because of condition 4, the linear map #L:V\to V# is symmetric if and only if #\dotprod{(L\vec{a}_1)}{ \vec{a}_2}=\dotprod{\vec{a}_1 }{(L\vec{a}_2)} #.
This theorem says that if a linear map is symmetric on a orthonormal basis, then this map is symmetric on the whole space. This is not very surprising, since every vector can be written as a linear combination of vectors in the orthonormal basis.
Here is the link between symmetric maps and symmetric 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.
Let #\alpha =\basis{\vec{a}_1,\ldots ,\vec{a}_n}# be an orthonormal basis for #V#. The matrix #L_\alpha# has as columns the #\alpha#-coordinates of the vectors #L(\vec{a}_1),L(\vec{a}_2),\ldots,L(\vec{a}_n)#. The #(i,j)#-entry of #L_\alpha# is the #i#-th coordinate of #L(\vec{a}_j)#. With a view to property 2 of orthonormal systems of vectors, this means: #\dotprod{\vec{a}_i}{(L(\vec{a}_j))}#. Likewise, the #(j,i)#-entry equals the #j#-th coordinate of #L(\vec{a}_i)#, so #\dotprod{(L(\vec{a}_i))}{\vec{a}_j}#. If #L# is symmetric, then these entries are equal. The matrix #L_\alpha# thus satisfies #L^\top_\alpha=L_\alpha#. That is to say, #L_\alpha# is symmetric.
Conversely, if #L_\alpha# is symmetric, then its #(i,j)#-entry is equal to its #(j,i)#-entry:
\[\dotprod{\vec{a}_i}{(L(\vec{a}_j))}=\dotprod{(L(\vec{a}_i))}{\vec{a}_j}=\dotprod{\vec{a}_j}{(L(\vec{a}_i))}\] Using the above theorem Characterizations of symmetry we conclude that #L# is symmetric.
As we saw in an example of the definition of a symmetric map, the orthogonal projection #P_W :V\rightarrow V# on a subspace #W# is symmetric. We can also establish this fact by use of the theorem. Choose an orthonormal basis #\vec{a}_1, \ldots,\vec{a}_n# of #V# such that #\vec{a}_1, \ldots ,\vec{a}_m# is an orthonormal basis of #W# (this is possible thanks to the Gram-Schmidt procedure). Each vector from the basis is an eigenvector: the first #m# vectors with eigenvalue #1# and the last #n-m# with eigenvalue #0#. Thus, the matrix of the map #P_W# with respect to above orthonormal basis is a diagonal matrix, and is therefore symmetric. By the theorem we may conclude that #P_W# is symmetric.
Write #n=\dim{V}#. Previously we saw that the map #\varphi: L(V,V)\to M_{n\times n}# which assigns to a linear map #L:V\to V# the matrix #L_\alpha#, is an isomorphism of vector spaces. This vector space has dimension #n^2#, the number of entries of an #(n\times n)#-matrix. On the basis of the first two properties of symmetric maps and the fact that the zero matrix is symmetric, the subset of #L(V,V)# of all symmetric linear maps #V\to V# is a linear subspace. This linear subspace is isomorphic (via the restriction of the isomorphism #\varphi#) to the image #\im{\varphi}#, which is the subspace of #M_{n\times n}# consisting of all symmetric matrices (this follows from the above theorem). The subspace has dimension #\frac12 n\cdot(n+1)#. This can be verified by use of the basis formed by the set #E_{ij}+E_{ji}# for all #i,j# with #1\le i\le j\le n#, where #E_{ij}# is the #(n\times n)#-matrix having as #(i,j)#-entry #1# and all other entries equal to #0#.
Consider the matrix #A = \matrix{a&b\\ c&d}#. The linear map #L_A:\mathbb{R}^2\to\mathbb{R}^2# determined by #A# is symmetric if and only if #A# is symmetric, that is, if and only if #b=c#.
Let #P_1# be the inner product space of all polynomials in #x# of degree at most #1# with orthonormal basis #\basis{1, x}#. Consider the linear map #L:P_1\to P_1# given by \[\eqs{ L(1) &=& -8 +3 x\\ L(x) &=& a+5 x}\] For what value of #a# is #L# symmetric?
#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
\[\begin{array}{rcl} \dotprod{(-8 + 3 x )}{{x}} &=& \dotprod{{1}}{(a+5 x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (-8)\cdot(\dotprod{1}{x})+3 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+5\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\
3 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\
\end{array}\] We conclude that #a = 3#.
Connection with symmetric matrices
