Boundedness of a matrix operator in a norm

Question

I would like to ask a simple question.

How do I show that a matrix is a bounded linear operator, for example this matrix $$A=\begin{bmatrix} t~~0\\ 0~~\frac{1}{\sqrt{t}} \end{bmatrix}$$ I know that the operator norm is given as $$||A||=\sup _{x\in H,x\neq 0} \frac{||Ax||}{||x||}$$

Which I believe is equivalent with $$||A||=\sup _{x\in H,||x||=1}||Ax||$$

I know that norm can be calculated using $$A^{T}A$$ and then finding the maximum eigenvalue, in this case I get , $t=\lambda,~~\lambda = \frac{1}{t}$ which blows up for $t\to \infty$ and therefore this operator isnt bounded?

Better call it $A_t$. It is bounded (as every matrix is, whatever the norm). Only the family $(A_t)_{t>0}$ is unbounded. — Anne Bauval, Jul 19 '24 at 19:51
Using the eigenvalue method I found that $\lambda = t$ and as $t\to \infty$ it blows up? — Vuk Stojiljkovic, Jul 19 '24 at 20:54
It is bounded for each $t$. In other words, no matter what $t$ is given to be, it is bounded. — Euclid, Jul 19 '24 at 21:31
It is the same for a number (a 1x1 matrix). For each $t\in\Bbb R$, $t$ is bounded. This bound is not uniform in $t$: the limit $t\to\infty$ is unbounded. — LL 3.14, Jul 19 '24 at 23:03

score 1 · Accepted Answer · answered Jul 20 '24 at 05:38

As explained in the comments, you are mixing up two different concepts: boundedness of a matrix and uniform boundedness of a family of matrices.

To illustrate this via the example \begin{align*} A:(0,\infty)&\to\mathbb R^{2\times 2}\\ t&\mapsto\begin{pmatrix} t&0\\0&\frac1{\sqrt t} \end{pmatrix} \end{align*} you used: When talking about boundedness of a matrix in the usual sense the parameter $t$ has to be fixed because the only allowed input for boundedness are Hilbert space vectors. In this case, for any fixed $t>0$ $$ \|A(t)\|_\infty=\max_{x\in\mathbb R^2}\sqrt{t^2x_1^2+\tfrac{x_2}t}=\max\{t,\tfrac1{\sqrt t}\}=\begin{cases} t&t\geq 1\\ \frac1{\sqrt t}&t\leq 1 \end{cases}\,. $$ Thus regardless of the (again, fixed!) input $t$, $\|A(t)\|_\infty<\infty$ so $A(t)\in\mathbb R^{2\times 2}$ is bounded. As Anne Bauval pointed out in her comment this is always true: every matrix—after all potential parameters have been fixed—is necessarily bounded.

The other concept you described is uniform boundedness: here one cares not about $\|A(t)\|_\infty$ but rather about the largest constant which bounds all matrices $A(t)$ at the same time, that is, $$ \sup_{t>0}\|A(t)\|_\infty\,. $$ As you pointed out this quantity in our example is $\infty$ so while $A(t)$ is bounded for every individual $t$—this is the common use of "boundedness"—the family $A(t)$ is unbounded.

Thank you so much for your comment. I have been busy recently with some other things and forgot the difference of these two concepts. I thought logically that it is similar to the concept of boundedness in classical analysis, so I thought how can $f(t)=t$ for example be bounded. I am very thankful to you for this. I just needed to understand the difference. If you can advise me some book concerning this matter, it would be great. In the books I read I did not encounter much of matrix family mappings. — Vuk Stojiljkovic, Jul 20 '24 at 08:50
@VukStojiljkovic One thing that comes to mind are the first two chapters in Kato's Perturbation Theory for Linear Operators. The whole premise of the book is to look at such families of matrices. However, he primarily looks at things like families of eigenvalues, eigenvectors, etc., mostly from the perspective of perturbation theory; from what I remember there is only little focus on things like uniform boundedness. So take this recommendation with a grain of salt. — Frederik vom Ende, Jul 20 '24 at 12:36

Boundedness of a matrix operator in a norm

1 Answers1