Connection with symmetric matrices

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 $i\lt j$ .

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 $\dotprod{(L\vec{a}_1)}{ \vec{a}_2}=\dotprod{\vec{a}_1 }{(L\vec{a}_2)}$ .

In words 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 $\eqs{ L(1) &=& 6 +6 x\\ L(x) &=& a+x}$ For what value of $a$ is $L$ symmetric?

$a =$ ${6}$

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
$\begin{array}{rcl} \dotprod{(6 + 6 x )}{{x}} &=& \dotprod{{1}}{(a+x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ (6)\cdot(\dotprod{1}{x})+6 \cdot(\dotprod{x}{x}) &=& a\cdot\dotprod{1}{1}+1\cdot(\dotprod{1}{x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\ 6 &=&a\\&&\phantom{xx}\color{blue}{\text{orthonormality of the basis}}\\ \end{array}$ We conclude that $a = 6$ .

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