For example, the Fredholm alternative says that if $Ax = 0$ has a non-trivial solution, then $Ax = b$ has a solution only if $b$ is orthogonal to to all solutions of $A^{T}x = 0$
Apparently this holds for linear operators as well as matricies, but I don't know how the transpose of a linear operator is defined
user71494
I have a question, if 1=0.999..., couldn't 1 mod 1 = 0.999 mod 1 = 1 = 0?
@Bitrex the transpose of a matrix is where you just flip the entries across the diagonals. More abstractly the dual of a map $f:V\to W$ is the map $f^*:W^*\to V^*$ given by $w^*\mapsto w^*\circ f$.
(note that with a fixed basis and dual basis, the matrix of the dual map is the transpose of the matrix of the original map)
@Bitrex If $A \colon X \to Y$, then $A^T \colon Y' \to X'$ is defined as $A^T(\lambda) = \lambda \circ A$. $X', Y'$ are the dual spaces (topological dual).
@Runemoro but then where are you getting (0.999... mod 1) = 0 from? you're not allowed to make up your own (false) rules in math. you may be interested in this related question.
user71494
@anon No, I said that 0.999... mod 1 = 0.999 and that 1 mod 1 = 0
@Bitrex in abstract algebra one learns that in the context of quotient rings, where both addition and multiplication is defined mod [blah], we must work with residues of integers. however we can still talk about the quotient group Q mod <1>=Z (i.e. the quotient group Q/Z)
@Runemoro okay, then where are you getting 1 = (0.999... mod 1) from? (I assuming you are using the notation [a mod b] as a binary operation returning an integer between 0 and b-1.)
@Runemoro are you using the programmer convention? if so then (0.999... mod 1) is equal to (1 mod 1) is equal to 0. if you use the mathematician convention then ... = -2 = -1 = 0 = 1 = 2 = ... (mod 1), i.e. every integer is congruent to every other integer mod 1.
@Bitrex Multiplication usually gets lost in these quotients by Z (i.e. in calculating mod 1) For instance .25*2 mod 1 is not the same as .25 mod 1 times 2 mod 1.
I think I need an example to help me get an intuitive grasp of what the transpose of a linear operator means. For example, lets say I have the linear operator $\frac{d}{dx}$ with some boundary conditions. How do I find its transpose?
@TedShifrin Well, the answer in my text is that the transpose of $\frac{d}{dx}$ with boundary conditions $y(0) = 0, y(1) = 1$ is the operator $-\frac{d}{dx}$ without any boundary conditions
@JohnWilson you asked me if I understood your question. on my screen your question has some square symbols in it. I am not familiar with that as a logical connective.
the adjoint $T^*$ is the unique map such that $\langle Tu,v\rangle=\langle u,T^*v\rangle$ for all $u,v$. even if the problem is using the word "transpose" it actually means the adjoint (which when talking about matrices is in fact the hermitian conjugate)
I am having trouble with a problem in the book I'm self-studying from. It says the following:
Show that $(P\to Q) \land (Q \to R) $ is equivalent to $(P \to R)$
$\land [(P \iff Q) \lor (R \iff Q)]$ by using logical connectives
I have dedicated so far a hefty amount of time on this problem...
I am having trouble with a problem in the book I'm self-studying from. It says the following:
Show that $(P\to Q) \land (Q \to R) $ is equivalent to $(P \to R)$
$\land [(P \iff Q) \lor (R \iff Q)]$ by using logical connectives
I have dedicated so far a hefty amount of time on this problem...
@TedShifrin Well, not one really. I finished reading chapter four of Spivak's book. Read the proof of Stoke's Theorem, impressed by the trivial nature of it. =)
@JohnWilson Passion. It's like loving someone, you don't really know why you love 'em, you just do. You may say "it's beautiful, nice, sweet" but there is a deeper reason that eludes you.
@Eugene In fact, I just finished a section of Edmund Landau's Elementary Number Theory where he proves Dirichlet's theorem. I remember you made fun of me studying from Apostol! >:(
Is there any generalization of the geometric mean that works as a norm in a normed space? The definition I have of a norm has that $||\vec{r}|| > 0$ when $\vec{r} \neq 0$, but the geometric mean is going to be zero if any component is zero
I am having trouble with a problem in the book I'm self-studying from. It says the following:
Show that $(P\to Q) \land (Q \to R) $ is equivalent to $(P \to R)$
$\land [(P \iff Q) \lor (R \iff Q)]$ by using logical connectives
I have dedicated so far a hefty amount of time on this problem...
