yes, these matrices are isomorphic under swapping of rows or columns @Agawa001, the same happen for the matrix $\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ and $\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ or $\left( \begin{array}{ccc} 0 & 0 \\ 0 & 1 \end{array} \right)$
indeed exist some papers about avoiding matrices of these kind... it maybe related with what I want but not completely
well i mention the discriminant = 0 so only one solution but i don't think saying the solution is {0} would really be explaining why its only one solution
@Dave discriminants are associated to polynomials, not equations (but people tend to mix them up quite a bit)
@Dave But anyway, discriminant 0 means there is at least one root with multiplicity at least 2, and since a quadratic can have at most two roots, that means there is only the one
The only real explanation you can give is that the value of x = 0 will, when substituted into the given equation will reduced to the true statement 0 = 0. Try it and prove it for yourself :-) @Dave
@skillpatrol the question says find the discriminant (which i got already as 0), what does this tell you about the number of solutions of the equation, explain briefly your answer.
Then it says, what does this tell you about the graph of y = 9x^2 -12x + 4
General solution of Transport equation (homogeneous): Method of Characteristics
$$u_t+cu_x=0 (\star)$$
We know that if $f: \mathbb{R} \to \mathbb{R}$ is differentiable then $u(x,t)=f(x-ct)$ is a solution of $(\star)$.
We will show that each solution of $(\star)$ is of the form $u(x,t)=f(x-ct)...
"A question: Why is this better than simply solving f=ma ?" - really, do you really want me to show that using acceleration (and the inherent problem of not having infinite small time steps in simulations) is not working to simulate orbits that should be repeatable?
@alkabary I don't know anything about Hungerford, but I learned character theory from Isaacs. It's very terse, if you're into that sort of thing. He has a book on character theory, and an Algebra for grad students book (called Algebra) that I believe does some representation theory.
Not to say it's better than Hungerford or anything, just an option you can look into
Let $V=C^1([a,b])$. If $J$ is a continuous functional for the norm $\|\cdot\|_\infty$ then it is continuous for the norm $\|\cdot\|_1:= ||y||_{\infty}+||y'||_{\infty}, y \in V$. But the converse is false.
In my book there is the hint that we can use the functional of arc length, in order to show...
Quick TeX question, I'm getting a syntax error which I've isolated to a line where the only non-standard text is my attempt to use the $\LaTeX$ symbol. It is returning a "you can't use \spacefactor in math mode." Is there some specific package I need to call to use the symbol?
(This is in an actual .tex document, not a MathJax page)
Show that the functional $J(y)=\int_a^b (\sin^3 x+y^2) dx$ is continuous in respect to the $||\cdot||_{\infty}$ norm, at any $y_0 \in C([a,b])$.
Let $y_0 \in C([a,b])$. Then for $y \in C([a,b])$ we have:
$$|J(y)-J(y_0)|=\left| \int_a^b (y^2-y_0^2) dx\right| \leq \int_a^b ||y-y_0||_{\infty} |y+y...
I just finished grading and typesetting responses to homework for a section of 'distance learning' students in a course I'm the grader for. In an effort to encourage them to type their work and not send in literal photographs of their work (you can even see the guy's thumb in every single shot), I'm posting responses in TeX and providing a link to miktex.org
@Bib a matrix, $A$, is defined to be idempotent iff $A=A^2$, yes? Is there some difference with "full idempotent"? The definition I'm reading is that an element, $e$, in a ring, $R$, is full idempotent iff $ReR=R$. That is the definition you are working with?
hmm, there are several examples of idempotent elements, the question of if they are full idempotent or not. Trivially, the identity satisfies the property.,
Another easy to prove one should be projection matrices, though I don't believe they would satisfy the "full" requirement.
Continue through the remaining multiplication to get what appears to be a full matrix. I'm starting to secondguess myself though, if the location of one zero forces too many conditions on the other entries of the matrix
yea... looking at the 3x3 case, the result is (sorry for mixing notation) $\begin{bmatrix} aa'&ab'&ac'\\ba'&bb'&bc'\\ ca'&cb'&cc'\end{bmatrix}$, but if we want a zero in the middle, that implies that either $b$ or $b'$ are zero, forcing either the entire second row or the entire second column to be zero
thus, we can't form $\begin{bmatrix}1&1&1\\1&0&1\\1&1&1\end{bmatrix}$
So... what does that leave us with? Here's a thought. If $e$ is invertible and is idempotent, then we have $e=e^2$ and that $1=e^{-1}e=e^{-1}e^2=e$
So, the only idempotent invertible matrix is the identity
suppose now that $e$ is not invertible, that implies that $det(e) = 0$
however, note that $\det(aeb) = \det(a)\cdot \det(e)\cdot \det(b) = 0$ since $\det(e)=0$
so, there is no way to have $ReR = R$ as an element on the righthandside if it were to be equal to $aeb$ must have determinant zero
thus, the only full idempotent matrix is the identity.
If someone wants to check my argument, I don't think I missed anything (I could rephrase it a bit better)
since $ReR\subsetneq R$ whenever $e$ is singular
Perhaps we should double check the definition that Lam uses for $E_{1,1}$ and what ring in particular he is referring to. If perhaps he intends that $E_{1,1}$ is infact the diagonal matrix of ones, or if perhaps the operation is not multiplication, then it could still be salvaged.
One of the easier (in my opinion) methods is to think of it via matrices and linear transformations @iluso You have $T\cdot\begin{bmatrix}u(n-1)\\u(n-2)\\u(n-3)\\\vdots\\u(n-8)\end{bmatrix} = \begin{bmatrix} u(n)\\u(n-1)\\u(n-2)\\\vdots\\u(n-7)\end{bmatrix}=\begin{bmatrix} 2u(n-1)-u(n-2)+2u(n-3)+\dots\\u(n-1)\\\vdots\end{bmatrix}$
You can then find a matrix representation for $T$. By diagonalizing the matrix, you can come up with a representation for $T^k$, and applying that to your initial vector $\begin{bmatrix} 56\\28\\14\\6\\2\\1\\1\\1\end{bmatrix}$ will give you a solution
In effect, the matrix argument is the same as the other approaches, the numbers which get raised to powers happen to be the eigenvalues of the matrix representation for $T$.
Unfortunatley, with a matrix that large, I wouldn't want to diagonalize it by hand, but computers can do so rather easily.