# The Unapologetic Mathematician

## Symmetric, Antisymmetric, and Hermitian Forms

The simplest structure we can look for in our bilinear forms is that they be symmetric, antisymmetric, or (if we’re working over the complex numbers) Hermitian. A symmetric form gives the same answer, an antisymmetric form negates the answer, and a Hermitian form conjugates the answer when we swap inputs. Thus if we have a symmetric form given by the linear operator $S$, an antisymmetric form given by the operator $A$, and a Hermitian form given by $H$, we can write

\displaystyle\begin{aligned}\langle w\rvert S\lvert v\rangle&=\langle v\rvert S\lvert w\rangle\\\langle w\rvert A\lvert v\rangle&=-\langle v\rvert A\lvert w\rangle\\\langle w\rvert H\lvert v\rangle&=\overline{\langle v\rvert H\lvert w\rangle}\end{aligned}

Each of these conditions an immediately be translated into a condition on the corresponding linear operator. We’ll flip over each of the terms on the left, using the symmetry of the inner product and the adjoint property. In the third line, though, we’ll use the conjugate-symmetry of the complex inner product.

\displaystyle\begin{aligned}\langle v\rvert S^*\lvert w\rangle&=\langle v\rvert S\lvert w\rangle\\\langle v\rvert A^*\lvert w\rangle&=-\langle v\rvert A\lvert w\rangle\\\overline{\langle v\rvert H^*\lvert w\rangle}&=\overline{\langle v\rvert H\lvert w\rangle}\end{aligned}

We can conjugate both sides of the last line to simplify it. Similarly, we can use linearity in the second line to rewrite

\displaystyle\begin{aligned}\langle v\rvert S^*\lvert w\rangle&=\langle v\rvert S\lvert w\rangle\\\langle v\rvert A^*\lvert w\rangle&=\langle v\rvert-A\lvert w\rangle\\\langle v\rvert H^*\lvert w\rangle&=\langle v\rvert H\lvert w\rangle\end{aligned}

Now in each line we have one operator on the left and another operator on the right, and these operators give rise to the same forms. I say that this means the operators themselves must be the same. To show this, consider the general case

$\displaystyle\langle v\rvert B_1\lvert w\rangle=\langle v\rvert B_2\lvert w\rangle$

Pulling both forms to one side and using linearity we find

$\displaystyle\langle v\rvert B_1-B_2\lvert w\rangle=0$

Now, if the difference $B_1-B_2$ is not the zero transformation, then there is some $w_0$ so that $B_1(w_0)-B_2(w_0)=v_0\neq0$. Then we can consider

$\displaystyle\langle v_0\rvert B_1-B_2\lvert w_0\rangle=\langle v_0\vert v_0\rangle=\lVert v_0\rVert^2\neq0$

And so we must have $B_1=B_2$.

In particular, this shows that if we have a symmetric form, it’s described by a self-adjoint transformation $S^*=S$. Hermitian forms are also described by self-adjoint transformations $H^*=H$. And antisymmetric forms are described by “skew-adjoint” transformations $A^*=-A$

So what’s the difference between a symmetric and a Hermitian form? It’s all in the fact that a symmetric form is based on a vector space with a symmetric inner product, while a Hermitian form is based on a complex vector space with a conjugate-symmetric inner product. The different properties of the two inner products account for the different ways that adjoint transformations behave.

July 10, 2009 Posted by | Algebra, Linear Algebra | 11 Comments