I'm a better learner through examples so I need help solving this one to better understand the operator norm.

We define first the norm of a matrix as $\|A\|_\infty = \sup_{1\leq i,j\leq n}|c_{ij}|$ .

Questions: Prove that $\Omega(A,B)=tr(AB)$ is continuous and calculate its operator norm.

I have no idea on how to proceed. I'm not trying to learn the solution. I just want to know the reasoning behind it. Am I supposed to find an inequality of some sort and bound the quantity? What's the method to proceed?

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    $\begingroup$ The operator norm (in the way you describe it) is defined for linear operators, but $\Omega$ is not linear (it's bilinear). $\endgroup$
    – MaoWao
    Commented Mar 19, 2021 at 20:55
  • $\begingroup$ Also, there's a notation issue where you define $\|A\|_{\infty}$ : there's two $=$ signs, I believe that's a typo. Second,what are the $c_{ij}$? Do not worry, we can give you the examples once you clarify these things. Mathematicians learn through examples, how to define properties and guess theorems, so you are taking the right route! $\endgroup$ Commented Mar 25, 2021 at 19:34
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    $\begingroup$ @TeresaLisbon yes that was a typo I edited the post! $c_{ij}$ are the elements of the matrix. $\endgroup$
    – Hijaw
    Commented Mar 25, 2021 at 19:48
  • $\begingroup$ @Hijaw Oh, that was quick, thanks! Another clarification : in the definition for $\|\Omega(A,B)\|$, on the RHS you have $\sup|tr(AB)|$, what is the supremum over? Another small issue : see, $A$ is a matrix so I can understand what $c_{ij}$ are and can get the answer. But what is $\|(A,B)\|_{\infty}$? I mean, how do I look at $(A,B)$ like a matrix? The best would be if you gave me a reference to the textbook you are reading (or online notes or video), and I can take the notation from there. $\endgroup$ Commented Mar 25, 2021 at 19:49
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    $\begingroup$ @TeresaLisbon that was my lousy attempt, has nothing to do with the exercice. I edited the question to remove it and just stick to what the exercice said. $\endgroup$
    – Hijaw
    Commented Mar 25, 2021 at 20:32

1 Answer 1


Since you said you only want to know how to approach this type of problem, I'll be a little vague here so you can fill in the details. First, to define the operator norm, as Teresa Lisbon mentioned in the comments, we can't really address this problem until we understand the spaces and norms we are talking about. You equipped the space $\mathbb{R}^{n\times n}$ with the $\ell^\infty$ norm. However, $\Omega$ is a mapping from $(\mathbb{R}^{n\times n})^2 \to \mathbb{R}$. To define the operator norm, we need to have a norm on the space $(\mathbb{R}^{n\times n})^2$. For the purposes of this problem, let $(A,B) = (a_{ij},b_{ij})_{i,j=1,\dots, n}$ and say that,

$$\|(A,B)\|_{\infty} = \sup_{i,j}\max\{|a_{ij}|,|b_{ij}|\}.$$

Then the definition of the operator norm is,

\begin{equation} \tag{1} \|\Omega\| = \sup_{\|(A,B)\|_{\infty} = 1} |\Omega(A,B)|. \end{equation}

From here on, I'm going to be a little vague. We can prove continuity directly. For any $\epsilon > 0$, introduce two perturbation matrices $\|(A_\epsilon,B_\epsilon)\|_\infty < \epsilon$. Then,

$$(A + A_\epsilon)(B+B_\epsilon) = AB + AB_\epsilon + A_\epsilon B + A_\epsilon B_\epsilon.$$

You can use this and the definition of $\Omega$ to directly bound the quantity $\|\Omega(A,B) - \Omega(A+A_\epsilon,B+B_\epsilon)|$ by some function of $\epsilon$. This would show that for any sequence $\{(A_n,B_n) \in (\mathbb{R}^{n\times n})^2\}$ such that $(A_n,B_n) \to (A,B)$, $\Omega(A_n,B_n) \to \Omega(A,B)$. Thus, $\Omega$ is continuous.

To compute the operator norm, we again consider equation (1). To compute $\|\Omega\|$, we want to show two things. First, we need a bound $M$ such that for all $(A,B)$ with $\|(A,B)\|_\infty=1$, $|\Omega(A,B)| \leq M$. Second, we need an example of matrices $\|(A,B)\|_\infty = 1$ such that $|\Omega(A,B)| = M$.

The easiest way to do this is to explicitly write out $\Omega$ in terms of the entries of $A$ and $B$. Note that,

$$\text{tr}(AB) = \sum_{k=1}^n (AB)_{k,k} = \sum_{k=1}^n \sum_{i=1}^n a_{k,i}b_{i,k}.\tag{2}$$

Then I'll leave this for you. Given that $|a_{i,j}|,|b_{i,j}| \leq 1$ for every $i,j \in \{1,\dots,n\}$, is there a simple choice of $M$ such that

$$\text{tr}(AB) \leq M?$$

Second, are there matrices $(A,B)$ with norm 1 such that $\text{tr}(AB) = M$?


Given that $-1\leq a_{ij},b_{ij}\leq 1$ for every $i,j$, is there a choice of entries $(a_{ij},b_{ij})_{i,j = 1,\dots,n}$ that clearly maximize equation (2)?

With this as an example, let's talk more generally. Suppose you have a normed vector space $X$ such that the set $\{x \in X: \|x\|_X= 1\}$ is compact and a continuous operator $\Omega: X \to \mathbb{R}$ and you want to compute the operator norm. Then try investigating what $\Omega(x)$ looks like and see if there is an $x \in X$ such that $\|x\|_X = 1$ and $|\Omega(x)| \geq |\Omega(x')|$ for any other $x' \in X$ such that $\|x'\|_X = 1$. If you can find that value of $x$, then $\|\Omega\| = |\Omega(x)|$. Note that the existence of this value $x$ follows from the continuity of $\Omega$ and the compactness of $\{\|x\|_X = 1\}$ in $X$.

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    $\begingroup$ Great stuff, +1. I am sure the OP will have loved the detailing and effort you have put in. $\endgroup$ Commented Mar 31, 2021 at 9:43

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