does integration by u substitution not work in all cases? for example if i have x*sin x dx , i don't know how to get just x dx to integrate since i will have cos x dx
If you see something of the form $f(x)g'(x)$, then integration by parts is a smart move.
In particular, it's usually what you want to do if you've got an integral like $x^n f(x)$. (Though in that case you usually are doing trig/hyperbolic trig/exponential functions, so taking antiderivatives is easy.)
How do I show $y'=f(x,y)=\frac{xy\cos(y)}{x^2+y^2},y(-1)=3/4$ has a unique solution which may be extended arbitrarily close to the boundary of the region $D=\{(x,y):-1\leq x\leq 1,y\in\Bbb{R},(x,y)\neq(0,0)\}$
@JessyunBourne induction - show you can apply elementary matrices in order to reduce A to a block matrix with blocks of size (n-1)x(n-1) and 1x1 (the lower block just being [1])
It's a lemma actually: Assume that at least one entry of $M$ is nonzero. Then using elementary transformations described above, we can transofmr $M$ to the form with zeroes in the first row and in the first column except $r_{11}$: $M = \begin{pmatrix}d & 0 & 0 & \cdots \\ 0 & r_{22} & r_{23} \cdots \\ 0 & r_{32} & r_{33} & \cdots \\ \cdots & \cdots & \cdots & \cdots& \cdots \end{pmatrix}$
But it doesn't say how to get $d = 1$, nor does it tell us how to take care of things when we want to keep our determinant $\pm 1$ or what to do when all the entries are integers.
@arctictern also, how many matrices do you need just to get it to that point up there?
@JessyunBourne well, $\det(M)=d\cdot\det(R)$, where $R$ is your lower block. If this $=1$, then $d=\pm1$. If $d=-1$ you can multiply by an elementary matrix to make it $+1$.
I'm saying that the resulting matrix is going to be $I_{n}$, so if the determinant changes signs along the way, it doesn't matter, because in the end it's going to be $+1$
I could just straight up apply the lemma if all I wanted to show was that $GL(n, \mathbb{Z})$ was finitely generated, but they want me to actually specifically say what the generating set is.
@arctictern yes. I got an extension on some assignments and the final for this course from last semester b/c my father was sick. I have until the end of the month
@arctictern no it uses elementary transformations.
@JessyunBourne Subtracting $t$ times row $2$ from row $1$ is just subtracting row $2$ from row $1$ a total of $t$ times, and subtracting one row from another is accomplished by multiplying by elementary matrices.
So you know what elementary matrices are, you know they are a finite generating set, but you don't know what the generating set is? You seem to be contradicting yourself.
I asked you: you already know what the generating set is, now you have to go about proving it, right?
Do you understand the effect of multiplying by elementary matrices on the left and right? (In other words, do you understand how these give us elementary transformations?)
I even get confused on that, because I don't really understand how to turn $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ using that language and using gcds
I know if I wanted to transpose rows, I'd multiply $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ on the left by the elementary matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
@Semiclassical I have shown that $Image(\phi_A)=\langle a_1, \ldots, a_n\rangle $, where $a_i$ stay for columns, that was pretty ok because of our conversation. Then I was given an $A=\begin{pmatrix} -2&1&3&2 \\ 1 & 0&-3&1 \\1&-1&0&-3\end{pmatrix}$ and have to find a basis $B$ of $kernel(\phi_A)$. I found out that kernel is the set of vectors $\{ \begin{pmatrix} -\lambda_4+3\lambda_3 \\ 3\lambda_2-4\lambda_4 \\ \lambda_3 \\ \lambda_4\end{pmatrix} \mid \lambda_4, \lambda_3 \in \mathbb{R}\}$.
If you think of the first two entries as (a,b), then row operations mean you can do things like (a,b)->(a,b+a) or ->(a,b-a) or (a+b,b) etc. In other words, you can do the Euclidean algorithm in the first two entries of the first column.
Once you get the first entry $\gcd(a_{11},a_{21})$ and the second entry $0$, you would do the Euclidean algorithm again, but this time using the first and third rows. This will end up with the third row having $0$ in the first column.
Well, at this point we've zeroed out the first row and column except $a_{11}$. Then $a_{11}$ must be $\pm1$. If $a_{11}=-1$ then multiply by ${\rm diag}(-1,1,\cdots,1)$. Otherwise we have proved the lemma. Then you want to use the lemma in an induction proof (or in other words, if you're describing the full algorithm, then yes repeat with the new lower block matrix, so we're zeroing out row&column a total of $n-1$ times).
@arctictern Okay, I'm going to try to write this up. How would an induction proof go for this though? The lemma would be my base case. Then, what's the induciton hypothesis?
What I'm getting at is that, for instance, $$\displaystyle \binom{\lambda_1}{\lambda_2}=\lambda_1 \binom{1}{0}+\lambda_2 \binom{0}{1}=\lambda_1 e_1+\lambda_2 e_2$$
But the fact that we're dealing with rows and columns of matrices is confusing.
The lemma isn't just showing hte case for $1\times 1$. It's showing that $a_{11} = 1$ and the rest of the row and column that $a_{11}$ is in are made up of zeros
It kind of seems then like $n-1$ is just the rest of the square
@Semiclassical so like $\lambda_4 \cdot \begin{pmatrix}-1\\-4\\0\\1\end{pmatrix}+\lambda_3 \cdot \begin{pmatrix}3\\3\\1\\0\end{pmatrix}$? And these two are my basis vectors? :)
If you don't think "reduce from nxn to (n-1)x(n-1) and the (n-1)x(n-1) case is the induction hypothesis" makes sense to you, then just describe the algorithm as you wanted to do so before I mentioned induction.
@kirill Not sure what you mean, but you certainly don't have to use those two vectors as a basis. Any basis of two vectors which generates the kernel will do.
(One has to be a bit careful with choosing a basis. For instance, do (1,0)^T and (2,0)^T generate the same subspace? Over $\mathbb{R}^2$ they do, but not over $\mathbb{Z}^2$!)
The lemma says you can multiply A by elementary matrices until it's a matrix A' with 1 in the upper left and an (n-1)x(n-1) matrix B in the lower right. By induction hypothesis, B is a product of elementary matrices. If we "frame" each of those elementary matrices (used to express B) by putting 0s to its left and on top of it and a 1 in the upper left, we have just written A' as a product of elementary matrices.
@Semiclassical got it. No, I thought, that as the dimension of my kernel is 2, I can just give two $e_i$ vektors and say that they build a basis. But, if everything were so simple, I could not get the purpose of my calculations with lambdas.
> Let $f:\Bbb R^3\to\Bbb R$ be given by$$f\begin{pmatrix}x\\y\\z\end{pmatrix}=z^2+4x^3z -6xyz+4y^3-3x^2y^2$$Is $M=f^{-1}(\{0\})$ a smooth surface (2-dimensional manifold)? If not, at what points does it fail to be so?
@JessyunBourne because $\det(A)=a_{11}\det($ lower block $)$
@10Replies the first two equations describe a line. parametrize the line, say as a function of t, then plug into third equation, try to solve for t (see which values of a make it possible)
@10Replies: You'll only have infinitely many solutions if $a^2-12=a+26=0$ (which can't happen). So can $a^2-12=0$ happen? If it doesn't, then I agree that there's a unique solution.
You're not dividing by zero. The equation says $0z = 23$, say. In other words, $0=23$. That cannot happen. But if the last equation says $0z = 0$, then the equation gives no information and you have just the first two equations.
It's still infinitely many solutions. When you have a free variable, it is in fact a parameter, but as it varies through all real numbers, you get infinitely many solutions.
If the coefficient of z is 0 and the constant on the right is not zero, you have 0=23, which is impossible. So there are NO solutions. Read what I wrote up above. I have to leave now. Good luck!
