The notion of symmetric map

Orthogonal and symmetric maps: Symmetric maps

The notion of symmetric map

In addition to orthogonal maps, symmetric maps form a second important class of linear maps on a real inner product space. Here we introduce these maps.

Let $V$ be a real inner product space. A linear map $L:V\rightarrow V$ is called symmetric if $\dotprod{L(\vec{x})}{\vec{y}}=\dotprod{\vec{x}}{L(\vec{y})}$ holds for all $\vec{x}, \vec{y}$ in $V$ .

Orthogonal projection The orthogonal projection $P_\ell:V\to V$ onto a straight line $\ell$ through the origin of an inner product space $V$ is symmetric. To see why, we choose a vector $\vec{a}$ in $V$ with ${\left\lVert \vec{a} \right\rVert} =1$ such that $\ell=\linspan{\vec{a}}$ . Then the orthogonal projection on $\ell$ is determined by ${P_\ell}(\vec{x})=(\dotprod{\vec{x}}{\vec{a}})\,\vec{a}$ For all $\vec{x}$ and $\vec{y}$ in $V$ we now have

$\begin{array}{rcl} \dotprod{( P_\ell(\vec{x}))}{\vec{y}}&=&\dotprod{\left((\dotprod{\vec{x}}{\vec{a}})\,\vec{a}\right)}{ \vec{y}}\\ &&\phantom{xx}\color{blue}{\text{mapping rule }P_\ell}\\ &=&(\dotprod{\vec{x}}{\vec{a}})\cdot (\dotprod{\vec{a} }{\vec{y}})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\ &=&(\dotprod{\vec{x}}{\vec{a}})\cdot (\dotprod{\vec{y} }{\vec{a}})\\&&\phantom{xx}\color{blue}{\text{symmetry of inner product}}\\ &=&\dotprod{\vec{x}}{\left((\dotprod{\vec{y}}{\vec{a})}\,\vec{a}\right)}\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\ &=& \dotprod{\vec{x}}{({P_\ell}(\vec{y}))}\\ &&\phantom{xx}\color{blue}{\text{mapping rule }P_\ell}\end{array}$ so $\dotprod{( P_\ell(\vec{x}))}{\vec{y}}=\dotprod{\vec{x}}{( P_\ell(\vec{y}))}$ . This shows that $P_\ell$ satisfies the definition of a symmetric map.

The above line $\ell$ is nothing but a $1$ -dimensional linear subspace of $V$ . In a similar way it can be shown that the orthogonal projection onto any linear subspace $W$ of $V$ is a symmetric linear mapping.

Let $L$ be the linear map ${\mathbb{R}^2\to\mathbb{R}^2}$ determined by the matrix $A=\matrix{a&b\\ c&d}$ Then $L$ is symmetric if and only if the matrix $A$ is symmetric; i.e., if and only if $b=c$ . Later we will prove a more general statement for all finite dimensions, but here we show that if $L$ is symmetric, then $b=c$ must hold:

$\begin{array}{rcl} b &=& \dotprod{\rv{1,0}}{\rv{b,d}} \\&&\phantom{xx}\color{blue}{\text{inner product calculation}}\\&=& \dotprod{\rv{1,0}}{L(\rv{0,1})}\\&&\phantom{xx}\color{blue}{\text{image is second column vector of }A}\\&=& \dotprod{L(\rv{1,0})}{\rv{0,1}}\\&&\phantom{xx}\color{blue}{\text{symmetry of }L}\\&=& \dotprod{\rv{a,c}}{\rv{0,1}}\\&&\phantom{xx}\color{blue}{\text{image is first column vector of }A}\\&=&c\\&&\phantom{xx}\color{blue}{\text{calculation of inner product}}\end{array}$

Let $V$ be an inner product space with basis $\basis{\vec{e}_1,\vec{e}_2}$ (so $V$ is $2$ -dimensional) and let $L:V\to V$ be a linear map. Then $L$ is symmetric if and only if $\dotprod{(L\vec{e}_1)}{\vec{e}_2} =\dotprod{\vec{e}_1}{(L\vec{e}_2)}$

This fact shows how the verification of symmetry is reduced to a calculation for a single pair of vectors. As we will see later, for higher dimensions, more calculations are needed, but still only for a finite number of vectors.

The statement is a consequence of the following chain of equivalent statements.

$\begin{array}{rcl}L \text{ is symmetric}& \Leftrightarrow &\text{ For all vectors }\vec{x},\vec{y}\text{ we have }\dotprod{(L\vec{x})}{\vec{y}} =\dotprod{\vec{x}}{(L\vec{y})}\\&&\phantom{xxx}\color{blue}{\text{definition of symmetry}}\\&\Leftrightarrow&\text{For all scalars }a,b,c,d\text{ we have }\\&&\phantom{xxxxx}\dotprod{(L(a\vec{e}_1+b\vec{e}_2))}{(c\vec{e}_1+d\vec{e}_2)} =\dotprod{(a\vec{e}_1+b\vec{e}_2)}{(L(c\vec{e}_1+d\vec{e}_2))}\\&&\phantom{xxx}\color{blue}{\vec{x} =a\vec{e}_1+b\vec{e}_2\text{ and }\vec{y}=c\vec{e}_1+d\vec{e}_2\text{ substituted}}\\ &\Leftrightarrow&\text{For all scalars }a,b,c,d\text{ we have }\\&&\phantom{xxxxx}\dotprod{(aL\vec{e}_1+bL\vec{e}_2)}{(c\vec{e}_1+d\vec{e}_2)} =\dotprod{(a\vec{e}_1+b\vec{e}_2)}{(cL\vec{e}_1+dL\vec{e}_2)}\\&&\phantom{xxx}\color{blue}{\text{ linearity of }L}\\&\Leftrightarrow&\text{For all scalars }a,b,c,d\text{ we have }\\&&\phantom{xxxxx}ac\,(\dotprod{L\vec{e}_1}{\vec{e}_1})+ad\,(\dotprod{L\vec{e}_1}{\vec{e}_2})+bc\,(\dotprod{L\vec{e}_2}{\vec{e}_1})+bd\,(\dotprod{L\vec{e}_2}{\vec{e}_2})\\&&\phantom{xxxxx}=ac\,(\dotprod{\vec{e}_1}{L\vec{e}_1})+ad\,(\dotprod{\vec{e}_1}{L\vec{e}_2})+bc\,(\dotprod{\vec{e}_2}{L\vec{e}_1})+bd\,(\dotprod{\vec{e}_2}{L\vec{e}_2})\\&&\phantom{xxx}\color{blue}{\text{linearity of the inner product }}\\&\Leftrightarrow&\text{For all scalars }a,b,c,d\text{ we have }\\&&\phantom{xxxxx}ad\,(\dotprod{L\vec{e}_1}{\vec{e}_2})+bc\,(\dotprod{L\vec{e}_2}{\vec{e}_1})=ad\,(\dotprod{\vec{e}_1}{L\vec{e}_2})+bc\,(\dotprod{\vec{e}_2}{L\vec{e}_1})\\&&\phantom{xxx}\color{blue}{\text{terms cancel since }\dotprod{L\vec{e}_j}{\vec{e}_j}=\dotprod{\vec{e}_j}{L\vec{e}_j}\text{ by symmetry of the inner product}}\\&\Leftrightarrow&\text{For all scalars }a,b,c,d\text{ we have }\\&&\phantom{xxxxx}ad\,(\dotprod{L\vec{e}_1}{\vec{e}_2})+bc\,(\dotprod{L\vec{e}_2}{\vec{e}_1})=ad\,(\dotprod{L\vec{e}_2}{\vec{e}_1})+bc\,(\dotprod{L\vec{e}_1}{\vec{e}_2})\\&&\phantom{xxx}\color{blue}{\dotprod{L\vec{e}_i}{\vec{e}_j}=\dotprod{\vec{e}_j}{L\vec{e}_i}\text{ by symmetry of the inner product}}\\ &\Leftrightarrow&\dotprod{L\vec{e}_1}{\vec{e}_2}=\dotprod{L\vec{e}_2}{\vec{e}_1}\\&&\phantom{xxx}\color{blue}{\text{the special case }a=d=1\text{ and }b=c=0\text{ suffices}} \end{array}$

Let $V$ be a finite-dimensional real inner product space and $L:V\rightarrow V$ a linear map. Then there is a unique linear map $L^\top:V\rightarrow V$ , the adjoint of $L$ , determined by $\dotprod{L(\vec{x})}{\vec{y}}=\dotprod{\vec{x}}{L^\top(\vec{y})}$ for all $\vec{x}, \vec{y}$ in $V$ .

Here is a proof of this statement. Let $\vec{y}$ be a vector of $V$ . Then the map sending $\vec{x}$ to the inner product $\dotprod{L(\vec{x})}{\vec{y}}$ is a linear map $V\to\mathbb{R}$ . Because $V$ is finite-dimensional, theorem The linear space of linear maps implies that there is a unique vector $\vec{a}$ such that $\dotprod{L(\vec{x})}{\vec{y}}=\dotprod{\vec{a}}{\vec{x}}$ for all $\vec{x}$ . The fact that $\vec{a}$ is uniquely determined by $\vec{y}$ means that there is a unique map $L^\top: V\to V$ , such that ${L^\top}(\vec{y})={\vec{a}}$ . Since the inner product is symmetric, this gives $\dotprod{L(\vec{x})}{\vec{y}}=\dotprod{\vec{x}}{L^\top(\vec{y})}$ for all $\vec{x}$ .

It remains to be proven that $L^\top$ is linear. For arbitrary scalars $\lambda$ , $\mu$ and vectors $\vec{y}$ , $\vec{z}$ we have

$\begin{array}{rcl}\dotprod{\vec{x}}{L^\top(\lambda\vec{y}+\mu\vec{z})}&=& \dotprod{(L(\vec{x}))}{(\lambda\vec{y}+\mu\vec{z})}\\&&\phantom{xx}\color{blue}{\text{definition of }L^\top}\\&=&\lambda \dotprod{(L(\vec{x}))}{\vec{y}}+\mu\dotprod{(L(\vec{x}))}{\vec{z}}\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product}}\\ &=&\lambda \dotprod{\vec{x}}{(L^\top(\vec{y}))}+\mu\dotprod{\vec{x}}{(L^\top(\vec{z}))}\\&&\phantom{xx}\color{blue}{\text{definition of }L^\top}\\&=&\dotprod{\vec{x}}{(\lambda L^\top(\vec{y})+\mu L^\top(\vec{z}))}\\&&\phantom{xx}\color{blue}{\text{bilinearity inner product}}\\\end{array}$

This implies $L^\top(\lambda\vec{y}+\mu\vec{z})=\lambda L^\top(\vec{y})+\mu L^\top(\vec{z})$ (because, as the above equality shows, the difference is perpendicular to $V$ ) which establishes the linearity of $L^\top$ .

The linear map $L$ is symmetric if and only if $L=L^\top$ . For, if $L=L^\top$ , then the uniqueness of the adjoint of $L$ implies that $L$ is symmetric; and if $L$ is symmetric, then $L$ satisfies the definition of $L^\top$ .

Here are some properties of symmetric maps.

Let $V$ be an inner product space, let $L$ and $M$ be symmetric linear maps $V\to V$ , and let $\lambda$ be a scalar.

$L+M$ is symmetric.
$\lambda\, L$ is symmetric.
If $L$ and $M$ commute (that is, $L\, M = M\, L$ ), then $L\, M$ is symmetric.
If $L$ is invertible, then $L^{-1}$ is also symmetric.

1. $L+M$ is symmetric: If $\vec{x}$ and $\vec{y}$ are vectors of $V$ , then $\begin{array}{rcl}\dotprod{((L+M)(\vec{x}))}{\vec{y}}&=&\dotprod{(L(\vec{x})+M(\vec{x}))}{\vec{y}}\\ &=&\dotprod{(L(\vec{x}))}{\vec{y}}+\dotprod{(M(\vec{x}))}{\vec{y}}\\&=&\dotprod{\vec{x}}{(L(\vec{y}))}+\dotprod{\vec{x}}{(M(\vec{y}))}\\&=&\dotprod{\vec{x}}{(L(\vec{y})+M(\vec{y}))}\\ &=&\dotprod{\vec{x}}{((L+M)\,(\vec{y}))}\end{array}$ 2. $\lambda\, L$ is symmetric: If $\vec{x}$ and $\vec{y}$ are vectors of $V$ , then $\begin{array}{rcl}\dotprod{((\lambda\, L)(\vec{x}))}{\vec{y}}&=&\dotprod{(\lambda\,(L(\vec{x})))}{\vec{y}}\\&=&\lambda\cdot\dotprod{(L(\vec{x}))}{\vec{y}}\\&=&\lambda\cdot\dotprod{\vec{x}}{(L(\vec{y}))}\\&=&\dotprod{\vec{x}}{(\lambda\,(L(\vec{y})))}\\&=&\dotprod{\vec{x}}{((\lambda\,L)\,(\vec{y}))}\end{array}$ 3. If $L$ and $M$ commute, then $L\, M$ is symmetric: If $\vec{x}$ and $\vec{y}$ are vectors of $V$ , then $\begin{array}{rcl}\dotprod{(L\,M(\vec{x}))}{\vec{y}} &=& \dotprod{(M(\vec{x}))}{(L(\vec{y}))}\\ &&\phantom{xx}\color{blue}{L\text{ is symmetric}}\\ &=& \dotprod{\vec{x}}{(M\,L(\vec{y}))}\\ &&\phantom{xx}\color{blue}{M\text{ is symmetric}}\\ &=& \dotprod{\vec{x}}{(L\,M(\vec{y}))}\\ &&\phantom{xx}\color{blue}{L\,M = M\, L}\end{array}$ 4. If $L$ is invertible, then $L^{-1}$ is also symmetric: If $\vec{x}$ and $\vec{y}$ are vectors of $V$ , then write $\vec{v} = L^{-1}(\vec{x})$ and $\vec{w} = L^{-1}(\vec{y})$ . Then we have $\dotprod{(L^{-1}(\vec{x}))}{\vec{y}}=\dotprod{\vec{v}}{(L(\vec{w}))}=\dotprod{((L(\vec{v}))}{\vec{w}}=\dotprod{\vec{x}}{(L^{-1}(\vec{y}))}$ .

Let $\vec{a}$ be a vector distinct from the zero vector of an inner product space $V$ and write $\ell=\linspan{\vec{a}}$ . Then, the reflection $S_{\vec{a}}$ about the plane $\linspan{\vec{a}}^\perp$ can be written as

$S_{\vec{a}} = I_V -2 P_{\ell}$

The identity map $I_V$ is clearly symmetric from the definition and $P_{\ell}$ is symmetric as we saw under the tab Orthogonal projection above. Therefore, statements 1 and 2 show that the reflection $S_{\vec{a}}$ is also symmetric.

The condition of statement 3 that $L$ and $M$ commute is necessary: Take $V = \mathbb{R}^2$ , let $L$ be determined by the matrix $\matrix{0&1\\ 1&1}$ and $M$ by the matrix $\matrix{0&2\\ 2&1}$ . Then $L\, M = \matrix{2&1\\ 2&3}$ so $\dotprod{(L\,M\,\rv{1,0})}{\rv{0,1}} =2\ne1= \dotprod{\rv{1,0}}{(L\,M\,\rv{0,1})}$

The symmetric linear maps $V\to V$ form a linear subspace of the vector space of all linear maps $V\to V$ . After all, the zero map is symmetric, and the first two statements show that the set of symmetric linear maps is closed under vector addition and scalar multiplication.

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 $\eqs{ L(1) &=& -9 -4 x\\ L(x) &=& a-3 x}$ For what integer $a$ is $L$ symmetric?

$a =$ ${-4}$

One of the requirements of symmetry on $L$ is $\dotprod{(L({1}))}{{x}}= \dotprod{{1}}{(L({x}))}$ . We work out this equation as follows
$\begin{array}{rcl} \dotprod{(-9 -4 x )}{{x}} &=& \dotprod{{1}}{(a-3 x)}\\&&\phantom{xx}\color{blue}{\text{mapping rule of }L\text{ used}}\\ -9 \cdot \dotprod{1}{{x}}-4 \cdot(\dotprod{x }{x}) &=& \dotprod{{1}}{a}-3\cdot(\dotprod{{1}}{ x})\\&&\phantom{xx}\color{blue}{\text{bilinearity of inner product }}\\ -4 &=&a\\&&\phantom{xx}\color{blue}{\dotprod{1}{x}=0\text{ and }\dotprod{x}{x}=1\text{ by orthonormality of the basis}}\\ \end{array}$ We conclude that $a = -4$ is the only possibility. Verifying that $\dotprod{(L(\alpha +\beta x))}{(\gamma+\delta x)} = \dotprod{(\alpha +\beta x)}{(L(\gamma+\delta x))}$ holds for arbitrary scalars $\alpha$ , $\beta$ , $\gamma$ , $\delta$ or application of the 2D criterion of the definition of symmetric maps shows that $a= -4$ indeed suffices for symmetry. Therefore, the answer is $a = -4$ .

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