Abstract Nonsense

Crushing one theorem at a time

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.


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{ }


\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}}



\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.




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


April 5, 2011 - Posted by | Algebra, Analysis, Linear Algebra | , , ,

1 Comment »

  1. […] Moreover, we’ll even show that if one gives each an inner product which is a multiple of the Hilbert-Schmidt inner product and the direct sum the usual inner product on direct sums that the isomorphism is also a unitary […]

    Pingback by Representation Theory: Decomposing the Group Algebra Into the Direct Sum of Matrix Algebras « Abstract Nonsense | April 6, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: