@KhallilBenyattou I can figure out the square root of any number in less than 10 seconds. What? You don’t believe me? Well, then, let’s try it with your phone number.
@KhallilBenyattou i'll send you a bunch when i get home. be wary. i grew up around engineers. we all have our minds in the gutter. we spend our days around nuts and bolts, lubricants, and sutff
I still don't get why no graph with a cut-vertex is possibly 1-factorable. I even have the proof infront of me and I still don't get the proof. Or why it's so.
@Studentmath tell me if at least I got the terminology right: a cut-vertex of a connected graph is a vertex such that if is removed the graph is no longer connected
Johnny had friends coming over the next morning. He set out three glasses of milk and went to bed. His friends came over and drank the glasses of milk. Later that afternoon, Johnny was planning to give each of his friends a cookie. How many cookies will Johnny need?
@DanielFischer No actually I did not. Do you know how to define the adjoint of an operator between Banach spaces? I only know a definition involving an inner product. (sorry was afk)
@MattN. The adjoint, or transpose, or dual of $T \colon E \to F$ is defined by $T^\ast(\lambda) = \lambda\circ T$. For Hilbert spaces, that corresponds to the Hilbert space adjoint via the Riesz map.
@DanielFischer I wonder why it's not mentioned here. Of course, most likely because Hermitian only refers to the Hilbert space case but I could find nowhere else on Wiki where they define it for Banach spaces.
"is formally similar to the defining properties of pairs of adjoint functors in category theory, and this is where adjoint functors got their name from." Hmmmmmmmmmmmmmmmmmmm, I'm almost sure adjoint operators were used and named before category theory came into being.
@mirgee That depends on the dimension and scalar field. One for dimension $0$. If the scalar field is $\mathbb{R}$, two if the dimension is $1$, if the dimension is larger, at least $2^{\aleph_0}$.
@mirgee If I have an orthogonal basis (they exist in finite dimension), multiplying all vectors in the basis by some $\lambda$ yields an infinite number of orthogonal basis
@mirgee $\{(1,0),(0,1)\}$ and $\{(i,0),(0,1)\}$ for example. You can always multiply any vector in an ON-basis with a scalar of modulus $1$ and get another ON-basis, so over $\mathbb{C}$, every inner product space of positive dimension has at least $2^{\aleph_0}$ ON-bases, since the unit circle has cardinality $2^{\aleph_0}$.
@mirgee Note further, that the image of an ON-basis under an orthogonal/unitary map is another ON-Basis, so you have a bijection between ON-bases and the orthogonal resp. unitary group of the space.
it's the graph with the smallest number of vertices such that each vertex has degree $5$ and the girth (the order of the smallest cycle subgraph) is $5$
@user127001 It's exactly as @Ted said, participation is the key to learning more. I stopped participating in a class and no doubt, it was my worst subject, statistics.
