Given a matrix $A$ over $\mathbb{R}$, define the operator norm as $\|A\|: = \sup\{\|A\mathbf{x}\| : \|\mathbf{x}\|=1\}$.  If $A$ is invertible, I realize that in general we have $\|A^{-1}\|\|A\|\geq 1$.   My question:  What is a concrete example where   this inequality is strict? I can't think of one.
62497
I am a beginner of funcional analysis. I have a simple question when I study this subject.  Let $L(X)$ denote the Banach algebra of all bounded linear operators on Banach space X, $T\in X$ is invertible, then $||T^{-1}||=||T||^{-1}$? Is this result correct?
64036
In a normed space $(E,\lVert \cdot\rVert)$ space we have the following inequality: $$\forall\, x,y\in E,\quad\|x\|^{2}-\|y\|^{2}\leq \lVert x-y\rVert\cdot \|x+y\|.$$ How can we prove it?
74589
Let $\mathrm{C}[0,1]$ be the space of continuous functions $[0,1]\rightarrow \mathbb{R}$ endowed with the norm $||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for $x(t)=t$ and $y(t)=1$). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties...
39439
Let $T$ be a linear operator on an inner product vector space $V$. I'd like to prove that if $$(Tx|x)=0 \quad \forall x \in V$$ then $T$ is the null operator. I can't figure out this proof using contradiction, do I have to choose an appropriate vector?
61040
Suppose we have an bounded linear operator A that operates from $L^2([a,b]) \mapsto L^2([a,b])$.  Now suppose that $A(f)(t) = tf(t)$.    Is A compact?  Edit: I know $A = A^*$ but I'm not really sure how to start on this.  It's not homework, just summer fun :D
37498
I wonder a generalization of Jensen's inequality: let $\mathbf{X,Y}$ be two positive definite matrices, can we obtain the following Jensen like inequality $$(1-\lambda)\mathbf{X}^{-1}+\lambda\mathbf{Y}^{-1} \succeq((1-\lambda)\mathbf{X}+\lambda\mathbf{Y})^{-1}$$ for any $\lambda \in (0,1)$, where the notation $\mathbf{A}\succeq \mathbf{B}$ represents matrix $(\mathbf{A-B})$ positive semi-definite.
17834
We define a symmetric, positive-definite matrix $A$ to be one such that $A = A^T$ and for $x \neq 0$, $x^TAx &gt; 0$.  If we have a norm $\|x\|_A = \sqrt{x^TAx}$, how can we show the triangle inequality?    That is, we want to show that $\|x + y\|_A \leq \|x\|_A + \|y\|_A$ as is the case for typical matrix norms.    $\|x+y\|_A^2 = (x+y)^TA(x+y) = (x^T + y^T)A(x+y) = x^TAx + x^TAy + y^TAx + y^TAy = \|x\|_A^2 + x^TAy + y^TAx + \|y\|_A^2$.    This is what I have so far, but I feel like I'm not really on the right track.  Ideas?
1188
It is a standard result that for a bounded self-adjoint operator $T$ on a complex Hilbert space $H$, we have $\|T\|=\sup_{\|x\|=1}|\langle Tx,x\rangle|:=M(T)$, where $M(T)$ is called the  of $T$. It seems that for any bounded operator on $H$, we have $||T||\leq 2M$. Why is that so?
23505
Let $-\infty\lt a\lt b \lt \infty$. Let $C[a,b]$ is space of continuous functions and a function $\|.\|:C[a,b]\to R$ is given by $\|f\|:=\int_a^b |f(x)|d(x)$  a). Show that $\|\cdot\|$ is a norm in $C[a,b]$  b). Show that $C[a,b]$ with this norm is not a Banach space.  Well the 1st question is kind of easy. Non-negativity, Homogenity, and Triangle inequality is easily proven. At the 2nd question I have a problem.   I know the $$C[0,1]$$ space with Norm $$\|f\| = \int_0^1|f(x)|d(x)$$ is not complete that means not Banach-Space. But how do I prove the generality?   Does anyone has a suggestion?  Thanks
57684
I got a positive definite matrix $B,$ that is, $V(x) = x^TBx &gt; 0$ for any vector $ x \neq 0.$  I want to show that $ \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2$ for any $x \neq 0,$ where $\lambda_\min$ and $ \lambda_\max$ are defined by $$ \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ and  $$ \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B  \rbrace$$  Any hint please?
88786
If $n \times n$ matrix $A \succeq 0$, and one $ n \times q$ column orthogonal matrix $U$, does this inequality hold? $$A- UU^{T} A UU^{T} \succeq 0$$
4413
I am trying to prove that $\|A\|=\sup_{||x||=1}|\langle x,Ax\rangle|$ for some selfadjoint bounded operator A on a Hilbertspace.  Can anyone give me a hint how to prove it.
96583
Let $B$ be a banach space and $T : B \to B$ be a bounded linear transformation.   If for identity transformation $I : B \to B$ , $||T-I||$&lt;1 , then $T$ is invertible.  || || is norm of transformation.
111157
Please help me with the following problem( give some hints or references):  Let $X$ be a Banach space and $B(X)$ be the algebra of bounded linear operators on $X$. Suppose that $A$ and $B$ are two operators in $B(X)$ such that for every $T \in B(X)$ we have $\sigma(AT)=\sigma(BT)$. Show that $A=B$.  Here $\sigma(A)$ denotes the spectrum of $A$, i.e., the the set of all complex numbers $\lambda$ such that $\lambda I − A$ has no inverse in $B(X)$.  Thanks in advance.
57003
Let $\| \cdot \|$ be a norm on $\mathbb{R}^n$. The associated dual norm, denoted $\| \cdot \|_*$ is defined as $\| z \|_* = \sup\{ z^{t} x : \| x \| &lt; 1 \}$.   Does someone know how to prove that the dual norm of the $\mathcal l_{p}$ norm is the $\mathcal l_{q}$ norm? I read about norms and it was stated without proof in a book.
23096
I know the title isn't very clear. The question is the following. Let $\|\cdot\|$ be a vector norm on $\mathbb{C}^n$. Define the induced norm on $\mathbb{C}^{n\times n}$ as: $$\|A\|_{op}=\max\limits_{x\in\mathbb{C}^n\smallsetminus\{0\}}\dfrac{\|Ax\|}{\|x\|}.$$ How do I prove that:   If $\|\cdot\|_{2}$ is the Euclidean norm, then: $$\|A\|_{2}=\rho^{\frac12}(A^{*}A),$$ with $\rho(A)$ the spectral radius of $A$ and $A^{*}$ the conjugate transpose of $A$? If $\|\cdot\|_{\infty}$ is the max norm ($\|x\|_{\infty}=\max|x_i|$, $x_i$ being the components of $x$) then: $$\|A\|_{\infty}=\max_{i=1,\dots,n}\left(\sum\limits_{j=1}^n|a_{ij}|\right)?$$ And that if $\|\cdot\|_{1}$ is the sum norm ($\|x\|_{1}=\sum_{i=1}^n|x_i|$, $x_i$ being, again, the components of $x$), then: $$\|A\|_{1}=\max\limits_{j=1,\dots,n}\left(\sum\limits_{i=1}^n|a_{ij}|\right)?$$
32065
Let $A$ be a unital $C^*$-algebra and $a\in A$ such that $r(a) &lt; 1$. Define b  = $(\sum_{n=0}^\infty (a^*)^n a^n)^{1/2}$. We can prove that $b\geq e$ and that $b$ is invertible. I want to show $\| b a b^{-1} \| &lt; 1$.   From the definition of $b$ we see that $a^* b^2 a = b^2-e$ and we know $r(bab^{-1}) = r(a) &lt;1$.   So it suffices to prove $r(b a b^{-1}) = \| b a b^{-1} \|$. It can follow from the fact $c = ba b^{-1}$ is a normal element... I don't know how to prove it (I have tried to compare $c^* c$ and $c c^*$...).    Context: The question appears in Murphy's book, page 74.   I have managed to prove the first part. The second part of the question is to prove $$r(a)= \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ It's easy to see $$r(a) \leq \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ But I can't prove $$r(a) \geq \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ If we have had $r(a) = \|b a b^{-1} \|$ then it was obvious... But this is not true, so how we can prove this inequality?