I have to show that $\sup_{n \in \mathbb{N}}|Tr(A_nB_n)| < \infty$. I know that $\sup_{n \in \mathbb{N}}||A_n||, \sup_{n \in \mathbb{N}}||B_n|| < \infty$, where $A_n,B_n$ are $2 \times 2$ random matrices and $ ||X|| = Tr(XX^T)^{1/2}$. Is there any inequality which I can use?
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$\begingroup$ I'm not sure of the exact matrix norm you're using. But can you use the definition of trace directly? I.e. $tr(AB) = \sum_{i=1}^n(AB)_{ii} = \sum_{i=1}^n \mathbf{A}_{row\,i}^\rm{T} \; \mathbf{B}_{col\,i} = \sum_{i=1}^n \sum_j (A)_{ij} (B)_{ji}$ $\endgroup$– TooToneJul 19, 2013 at 17:58
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2$\begingroup$ Then they have finite entries and obviously the trace of their product is finite! You can obtain a much stronger result, though, using the Cauchy-Schwarz Inequality. (Whether the entries are random or not is immaterial.) $\endgroup$– whuber ♦Jul 19, 2013 at 18:36
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1$\begingroup$ @TooTone: en.wikipedia.org/wiki/Supremum. Kolibris: the inequality applies directly by viewing the matrices as $4$-dimensional vectors. There's little more to be said; the rest is an application of the definition of supremum. $\endgroup$– whuber ♦Jul 19, 2013 at 18:51
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1$\begingroup$ @TooTone Another approach which you might like is tantamount to the Holder Inequality: from the axiomatic relation $||A_n-B_n||^2\ge 0$ it is straightforward to deduce $2Tr(A_nB_n')\le||A_n||^2+||B_n'||^2$, whence upper bounds $\alpha$ for $\{||A_n||\}$ and $\beta$ for $\{||B_n'||\}$ give an upper bound of $(\alpha^2+\beta^2)/2$ for $\{Tr(A_nB_n)\}$, etc. $\endgroup$– whuber ♦Jul 19, 2013 at 19:25
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1$\begingroup$ @TooTone The supremum is applied here to sequences of numbers, not matrices: the numbers are $Tr(A_nB_n)$, $||A_n||$, and $||B_n||.$ It might help to interpret the question in words as "when there is an upper bound to the norms of the $A_n$ and an upper bound to the norms of the $B_n$, then there is also an upper bound to the norms of the inner products $Tr(A_nB_n)$. So, if all the $A$'s can't get too big and all the $B$'s can't get too big, then there's a (finite) limit to how big all their inner products can get." That should make the truth of this assertion obvious. $\endgroup$– whuber ♦Jul 19, 2013 at 23:31
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1 Answer
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Following @whuber's suggestion, starting with
$$\begin{eqnarray*} A &=& \left(\begin{array}{rrrr} a & b \\ c & d \end{array}\right)\\ P &=& \left(\begin{array}{rrrr} p & q \\ r & s \end{array}\right)\\ \end{eqnarray*}_.$$
then the matrix product is
$$\begin{eqnarray*} AP^T &=& \left(\begin{array}{rrrr} a & b \\ c & d \end{array}\right) \left(\begin{array}{rrrr} p & r \\ q & s \end{array}\right)\\ &=& \left(\begin{array}{llll} ap+bq & \ldots \\ \ldots & cr + ds \end{array}\right)\\ \end{eqnarray*}_.$$
The norm of a matrix $$\begin{eqnarray*} ||A|| &=& Tr(AA^T)^{1/2} \\ &=& (a^2 + b^2 + c^2 + d^2)^{1/2} \end{eqnarray*}_,$$
is just the Euclidian norm of the matrix treated as a vector of its coefficients $\mathbf{a} = (a, b, c, d)^T$.
The trace of the product $$\begin{eqnarray*} Tr(AP^T) &=& ap + bq + cr + ds \end{eqnarray*},$$
is just the dot product (inner product) of the matrices treated as vectors of their coefficients.
The Cauchy-Schwarz inequality states that
$$ |\mathbf{x}\cdot\mathbf{y}| \le ||\mathbf{x}||\,||\mathbf{y}||,$$
where the vector norms on the right are Euclidian. This inequality can then be applied to the vectors $\mathbf{a}$ and $\mathbf{p}$ of coefficients of the matrices $A$ and $P$.
$$ |Tr(AP^T)| \le ||A||\,||P|| $$
We are given $\sup_{n\in N}||A_n|| < \infty$, and $\sup_{n\in N}||B_n|| < \infty$. Hence
$$ \sup_{n\in N}|Tr(AP^T)| \le \sup_{n\in N}(||A||\,||P||) < \infty, \;\;\mbox{and}$$ $$\sup_{n\in N}|Tr(AP^T)| < \infty .$$
However this is not the form required in the question: it gives an inequality for the trace of the product $AP^T$ rather than $AP$. But $||X||=||X^T||$, which follows from the fact that for Euclidian vector norms $||\mathbf{x}^T|| = ||\mathbf{x}||$, or equivalently, for any matrix $X$, $Tr(XX^T)=Tr(X^TX)$. Setting $B=P^T$ $$\sup_{n\in N}|Tr(AB)| < \infty .$$
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1$\begingroup$ I think that you can start with $AP$ instead of $AP^T$. In that case there is no problems with a transpose. whuber: an idea to vectorize matrices is very nice. $\endgroup$– KolibrisJul 20, 2013 at 13:31
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$\begingroup$ (On the spelling of Cauchy-Schwarz: it's Hermann Amandus Schwarz, rather than Schwartz.) $\endgroup$ Oct 3, 2015 at 0:14