For example, if $A = \begin{bmatrix}2&3\\0&1\end{bmatrix}$ then the characteristic polynomial of $A$ is: $ \begin{vmatrix}x-2&-3\\0&x-1\end{vmatrix} = (x - 2)(x - 1)$.
If the last 80% have EXACTLY one 1 per row, you should be able to re-arrange the rows to get a matrix in block form like so: $M = \begin{bmatrix}A&B\\0&I\end{bmatrix}$
Suppose that we have a uniform distributed random variable in $[0,n]$.
We have the following:
$$\max \{n-q,q-1\}=\left\{\begin{matrix}
q-1, q > \lfloor \frac{n}{2}\rfloor & \to q-1> \lceil\frac{n}{2} \rceil \\
n-q, q \leq \lceil \frac{n}{2} \rceil& \to n-q \geq n- \lceil \frac{n}{2}\rceil=\lf...
so if I search for the minimal polynomial, I can get the eigenvalues, and if I can get those, I can convert to JNF, and if I can do that, I can get the charpol?
Well, sort of. With a 3000-size matrix I suspect you'll have some multiplicities.
I mean, I think you can save some effort, by trying to "trap" information in the upper-left, and lower-right, but since I don't know what your matrix MEANS, I don't know specifically, how to tell you to get there.
Let $E = \{f \in \mathbb{P}_3 : f(0) = f(1)\}$, where $\mathbb{P}_3$ is the set of all polynomials with $\deg(f) \leq 3$ with coefficients in $\mathbb{R}$.
I have already shown that $E$ is a subspace of $\mathbb{P}_3$.
How do I find a basis for $E$ and $\dim(E)$?
Here's what I recall:
A basis for some vector space $V$ is a set of vectors in $V$ such that the vectors span $V$ and are linearly independent.
Now do I remember anything beyond that? Other than really basic standard bases, nothing really.
There are two versions of the Knapsack problem, the integer and the fractional one.
The difference between the integer and the fractional version of the Knapsack problem is the following: At the integer version we want to pick each item either fully or we don't pick it. At the fractional version...
@Clarinetist A common basis for $\mathbb{P}_3$ is $\{1,x,x^2,x^3\}$. You have also that $f(0)=f(1)$ for all $f\in E$. So, for $f(x) = a + bx + cx^2 + dx^3$, you have $f(0) = a = f(1) = a + b + c + d$ and so $b+c+d = 0$
@MaryStar I'm not very familiar with the problem you are working on, sorry
@Clarinetist so, $a$ is free to be whatever. If we allow $b$ and $c$ to be free, then you have that $d$ will be forced to be $d=-b-c$
@Clarinetist To write out a basis, it helps to write each basis element involving only a single free variable wherever it appears. As such, one choice of basis for $E$ could be $\{1, x - x^3, x^2 - x^3\}$
it might help as well to think of it as $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0\end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 0 \\ -1\end{bmatrix}$, etc...
Let $A$ and $B$ be nonempty and closed in $\Bbb C$.
Assume $A\cup B$ is not open, then there is a limit point $p$, not in $A\cup B$, that must be in $\overline{A}$ or $\overline{B}$, which means that
$$p\in\overline{A}\cup \overline{B}$$, but $A$ and $B$ are closed and hence $A=\overline{A}$ an...
E.g. where is it non-rigorous?
Maybe just edit it with a \color{red}{blahblahhhhhhhhh} field?
good I am planning to finish dummit over the summer and read some algebra advanced reading so it would be cool to discuss with you aswell over the summer.
I see. Well, I've only studied more or less what requires to understand Galois theory, so I doubt I would be helpful with the other chapters. I do look forward to study a bit homological algebra and group cohomology from Dummit-Foote though.
like if I have 2/3x that is normally represented as a float, but now since I am working mod p, I have (2 * x * (inverse of 3 mod p)) mod p, does this make sense?
So a professor teaches his class the Mean Value Theorem using an unusual method. Later, another teacher asks him, "Why did you prove the Mean Value Theorem in such a weird way?" He replies, "That, my friend, is how I Rolle."
@meer2kat What you need to show is: if $a = a'$ mod $n$, and $b = b'$ mod $n$, that: $ab = a'b'$ mod $n$
Here's why: the goal of multiplication mod $n$, is to set: $(a\text{ (mod }n))(b\text{ (mod }n)) = ab\text{ (mod }n)$.
But $a$ mod $n$ isn't "a single number", it's an entire equivalence class of numbers.
And we need to ensure that our calculation doesn't depend on "which" representative $a$ we pick.
But solely on the equivalence class.
So, if we're working mod 5, for example, we might have 4*3 = 2. But if we chose 14, and -7 instead, we should get something "equivalent" to 2 mod 5, when we multiply.
Which we do, we get -98, which is -100 + 2 = 2 mod 5.
When you read a book and feel it's a bit tough and even painful... and you just start staring at the page for 20 minutes without thinking.... what would you do?
@DavidWheeler you know I am currently reading the proof of cauchy's theorem for abelian groups I mean it is under stable but how come did cauchy see the intuition of that theorem I guess he must have tried alot of examples
@YilinWang Usually one has the basic "metric" criterion: for any $\epsilon > 0$ there is a $\delta$-ball $B_{\delta}(a)$ such that $f(B_{\delta}(a)) \subset B_{\epsilon}(f(a))$.
@DavidWheeler for this question If G is a not abelian group of order 15, prove that Z(G) = {e}. I proved it using the fact G/Z(G) is cyclic then G must be abelian and using lagrange to rule out 3 and 5.
However can it be proved using class equations ?
I was thinking
I mean from class equation we have |G| = [G: Z(G)] + [G: C($a_1$)] + ... + [G: C($a_r$)]
@JasperLoy The idea in the book is to make things clear for anyone interested in those integrals, series and limits. But who says anything about geniuses? Well, we're all just normal people (with some passion).
Because of the lack of sleep I have headache and no power in hands. Maybe I should take a nap, but later, not now.
Suppose that we have a uniform distributed random variable in $[0,n]$.
We have the following:
$$\max \{n-q,q-1\}=\left\{\begin{matrix}
q-1, q > \lfloor \frac{n}{2}\rfloor & \to q-1> \lceil\frac{n}{2} \rceil \\
n-q, q \leq \lceil \frac{n}{2} \rceil& \to n-q \geq n- \lceil \frac{n}{2}\rceil=\lf...
Let $A$ and $B$ be nonempty and closed in $\Bbb C$.
Assume $A\cup B$ is not open, then there is a limit point $p$, not in $A\cup B$, that must be in $\overline{A}$ or $\overline{B}$, which means that
$$p\in\overline{A}\cup \overline{B}$$, but $A$ and $B$ are closed and hence $A=\overline{A}$ an...
