# Abstract Nonsense

## The Hilbert-Schmidt Inner Product on Complex Matrix Algebras

Point of post: In this post we discuss a natural way to define an inner product on a complex matrix algebra of the form $\text{Mat}_n\left(\mathbb{C}\right)$ and describe some of the properties.

Motivation

Of course we know that the space of $n\times n$ of matrices over $\mathbb{C}$, $\text{Mat}_n\left(\mathbb{C}\right)$,  is an associative unital algebra with the usual definitions of scalar multiplication, matrix addition, and matrix multiplication. It may then seem fruitful to ask what kinds of inner products one can put on it. In this post we shall define a natural one, namely $\left\langle A,B\right\rangle_{\text{HS}}=\text{tr}\left(AB^\ast\right)$ called the Hilbert-Schmidt inner product and derive some interesting results about it.

Hilbert-Schmidt Inner Product

Let, as usual, $\text{Mat}_n\left(\mathbb{C}\right)$ denote the space of all $n\times n$ square matrices with the usual operations. We define the map

$\text{ }$

$\langle\cdot,\cdot\rangle_{\text{HS}}:\text{Mat}_n\left(\mathbb{C}\right)\times\text{Mat}_n\left(\mathbb{C}\right)\to\mathbb{C}$

$\text{ }$

by $\left\langle A,B\right\rangle_{\text{HS}}=\text{tr}\left(AB^\ast\right)$  where as usual $A^\ast$ is the conjugate transpose (i.e. if $A=[a_{i,j}]$ then $A^\ast=[b_{i,j}]$ where $b_{i,j}=\overline{a_{j,i}}$).  This is called the Hilbert-Schmidt inner product on $\text{Mat}_n\left(\mathbb{C}\right)$. What we’d like to claim is that $\langle\cdot,\cdot\rangle_{\text{HS}}$ is an inner product on $\text{Mat}_n\left(\mathbb{C}\right)$. Some of the properties, such as linearity in the first entry, and conjugate symmetry are obvious. The fact that it is positive definite is not. For this we consider the following lemma:

$\text{ }$

Lemma: Let $A=[a_{i,j}]\in\text{Mat}_n\left(\mathbb{C}\right)$ then $\displaystyle \text{tr}\left(AA^\ast\right)=\sum_{i,j=1}^{n}|a_{i,j}|^2$.

Proof: By definition, if we let $A^\ast=[b_{i,j}]$ we have that the general term $(r,s)^{\text{th}}$ term of $AA^\ast$ is

$\text{ }$

$\displaystyle c_{r,s}=\sum_{m=1}^{n}a_{r,m}b_{r,s}=\sum_{m=1}^{n}a_{r,m}\overline{a_{s,m}}$

Thus,

\displaystyle \begin{aligned}\text{tr}\left(AA^\ast\right) &=\sum_{r=1}^{n}c_{r,r}\\ &=\sum_{r=1}^{n}\sum_{m=1}^{n}a_{r,m}\overline{a_{r,m}}\\ &=\sum_{r=1}^{n}\sum_{m=1}^{n}\left|a_{r,m}\right|^2\\ &=\sum_{r,m=1}^n \left|a_{r,m}\right|^2\end{aligned}

from where the conclusion follows. $\blacksquare$

$\text{ }$

$\text{ }$

With this we are now ready to prove the main result of this post. Namely:

$\text{ }$

Theorem: The Hilbert-Schmidt inner product $\left\langle A,B\right\rangle_{\text{HS}}=\text{tr}\left(AB^\ast\right)$ is an inner product $\text{Mat}_n\left(\mathbb{C}\right)$.

Proof: Clearly $\langle\cdot,\cdot\rangle$ is linear in the first column since for every $a,b\in\mathbb{C}$ and  $A,B,C\in\text{Mat}_n\left(\mathbb{C}\right)$ one has that

$\text{ }$

\displaystyle \begin{aligned}\left\langle aA+bB,C\right\rangle_{\text{HS}} &= \text{tr}\left(\left(aA+bB\right)C^\ast\right)\\ &= \text{tr}\left(aAC^\ast+bBC^\ast\right)\\ &= a\text{tr}\left(AC^\ast\right)+b\text{tr}\left(BC^\ast\right)\\ &= a\left\langle A,C\right\rangle_{\text{HS}}+b\left\langle B,C\right\rangle_{\text{HS}}\end{aligned}

The fact that $\langle\cdot,\cdot\rangle$ is conjugate symmetric is clear since $\text{tr}$ is transpose invariant and distributive over conjugate so that

$\text{ }$

$\left\langle A,B\right\rangle_{\text{HS}}=\text{tr}\left(AB^\ast\right)=\text{tr}\left(\left(BA^\ast\right)^\ast\right)=\overline{\text{tr}\left(BA^\ast\right)}=\overline{\left\langle B,A\right\rangle_{\text{HS}}}$

Lastly, we see from our lemma that $\langle\cdot,\cdot\rangle$ is positive semi-definite since if $A=[a_{i,j}]\in\text{Mat}_n\left(\mathbb{C}\right)$

$\displaystyle \left\langle A,A\right\rangle_{\text{HS}}=\text{tr}\left(AA^\ast\right)=\sum_{i,j=1}^{n}\left|a_{i,j}\right|^2$

from where positive semi-definitness follows. Thus, the function $\langle\cdot,\cdot\rangle_{\text{HS}}$ satisfies all the conditions to be a complex inner product and thus the conclusion follows. $\blacksquare$

$\text{ }$

From this we can define the Hilbert-Schmidt norm, denoted $\|\cdot\|_{\text{HS}}$$\text{Mat}_n\left(\mathbb{C}\right)$ by $\left\|A\right\|_{\text{HS}}=\sqrt{\left\langle A,A\right\rangle_{\text{HS}}}$. We claim that this is a matrix norm (i.e. it’s a norm on $\text{Mat}_n\left(\mathbb{C}\right)$ in the usual sense and it also satisfies $\left\|AB\right\|_{\text{HS}}\leqslant \left\|A\right\|_{\text{HS}}\left\|B\right\|_{\text{HS}}$). Indeed, this follows immediately from the Cauchy-Schwarz inequality.

References:

1. Horn, Roger A., and Charles R. Johnson. Matrix Analysis. Cambridge [u.a.: Cambridge Univ., 2006. Print.