@Yai0Phah I'm afraid I haven't seen homotopy pushouts yet, so I can't quite grasp what you're saying (and I only read about the homotopy extension property this week, so I'm not entirely comfortable there yet either).
@ShaVuklia You don't have to "completely" grasp, but it seems better to be aware of this point of view earlier (even without much comprehension). You could look at arxiv.org/abs/1508.01942 for an introduction to the model structure on topological spaces to grasp some basic ideas (it is very self-contained and well written), and come back to it from time to time.
@TedShifrin i agree on this, but i wonder what the threshold value here is. the condition on reality of roots for a quadratic is easy, and for a cubic i feel lke there should be something. but quartic?
there are some galois theoretic results for polynomials with all real roots. and algorithms for determining if they are solvable in terms of field operations with real root extraction. not subtle stuff and you do run into obstructions.
I was just thinking, for the proof that there exists a basis $B$ such that the matrix $[T]_B$ is upper triangular where $T: V \rightarrow V$ with $V$ a vector space on $\mathbb{C}$. Can I not just use the proof for the triangularisation theorem and then use gram-schmidt on the basis for the quotient space???
yeah, it's theoretically clear that such a basis exists once you know that a non-orthonormal basis exist, because GS respects the set of subspaces spanned by the vectors even as it orthogonalizes and normalizes them. i thought the point of the exercise was actually to compute such a thing (where short word steps like "there is an eigenvector," and "there is v not in the span of u, w," etc. in general results get replaced with calculation)
govind: it would preserve upper triangularity in your setting. the matrix of T with respect to {e_1, e_2, ..., e_n} is upper triangular if and only if T(e_j) is in span {e_i: i <= j} for all j. if you gram schmidt a basis {e_1, .., e_n} into {f_1, ..., f_n}, span {e_i: i <= j} = span {f_i: i <= j} for all j because of how the algorithm works. so upper triangularity is preserved.
You can also think of that in terms of the change-of-basis theorem. When you GS, you get an upper-triangular matrix $P$, so $PBP^{-1}$ is still the product of upper-triangular matrices.
$\begin{bmatrix} \lambda & Z\\ 0 & A \end{bmatrix}$, If $A$ is $n-1 \times n-1$ and upper triangular with an orthonormal basis $\bar{B}$, and I wanted to add an eigenvector to complete the upper-left entry, would then doing GS on the basis $B = \{v\} \cup \bar{B}$, give me the orthonormal basis I need to get the upper triangular matrix above? As GS preserves upper-triangularity?
govind something worth thinking about is that in general Z is going to be nonzero. i.e. even if you split off the eigenvector there is nothing stopping T from mapping stuff orthogonal to it to stuff that isn't orthogonal to it. so analyzing A is not quite the same thing as analyzing T on a subspace. this doesn't break the logic of a properly structured proof, but should affect how you think about it.
there's often some subtlety in finding the right inductive hypothesis.
i suppose this is where ted's geometry comes in. it's possible to read through a book like axler and never notice all the potentially wrong choices that he isn't making in his inductive arguments.
@TedShifrin Lol Ted, I'm not trying to be obnoxious, but phrasing things categorically is the only way that I can keep track of identifications, which has cleared up so much confusion from my undergraduate years
I am trying to apply Duhamel's Principle starting with the very simple equation $P'(t) - \frac{1}{10}P(t)\ =\ \pi,\ P(0)\ =\ 0$. I started by solving $Q'(t) - \frac{1}{10}Q(t)\ =\ 0,\ Q(0)\ =\ \pi$, whose solution is $Q(t)\ =\ \pi e^{\frac{1}{10} t}$. How exactly would I set up the integral to solve for $P(t)$ using this $Q(t)$, or have I messed up the procedure somehow by using the "same notion of time" (the variable $t$) for $P$ and $Q$?
So assuming that the result holds for $n-1 \times n-1$ matrices and then decomposing the matrix into the eigenvector and quotient map and applying it is not correct?
@Govind Start with the eigenvector and take the orthogonal complement. What does your inductive hypothesis now tell you?
You cannot lift the quotient back to the original vector space to guarantee orthogonality.
You will end up with below-diagonal garbage.
@Govind The argument I suggest is in most books (probably all) — including my two that treat linear algebra.
@user10478 I've forgotten Duhamel. Don't you just convolve the $g$ on the RHS with the solution of the homogeneous equation? I guess I can check that for myself.
So the orthogonal complement will give me a basis of $n-1$ vectors $B^{\bot}$. Then taking $B = \{v\} \cup B^{\bot}$, and checking $[T]_B$ will give me my orthogonality, then by the induction hypothesis the $n-1 \times n-1$ bottom right sub-block will have an orthonormal basis $B*$, which is orthogonal to my eigenvector. So taking the final basis to be $\{\frac{v}{||v||}\} \cup B^*$ is my final orthonormal basis?
@TedShifrin There is a way to solve state-free ODEs such as my ODE in $P$ by convolving the impulse response with the forcing term. I did just solve the same problem by this method, and part of what I am trying to figure out is how Duhamel's Principle relates to this approach, since they seem procedurally similar but not exact.
@TedShifrin Duhamel's is where you start with an ODE or PDE with a nonzero forcing term but a zero initial condition, and move the forcing term to the initial condition, then take some integral of the resulting solution to get a solution to the original problem.
I might have to take a derivative too before the integral.
because by the induction hypothesis the bottom right sub-block has that basis which makes it upper triangular where each element is in $\mathbb{C}^{n-1}$, then taking the union of that basis with our normalized n-dim eigenvector will make sure the whole matrix is upper triangular. But how do I convert my basis elements in $\mathbb{C}^{n-1}$, to ones in $\mathbb{C}^n$
Can we be a bit more precise, @user10478. What is the "some integral"? Is it actually a convolution? If not, these are unrelated.
@Govind: You've just done an argument by intimidation. WHY is the net matrix upper-triangular? It's obvious, but you have to explain it carefully. ... Your $\Bbb C^{n-1}$ is sitting as a subspace of $\Bbb C^n$ to start with — it's the orthogonal complement of our one-dimensional subspace.
@TedShifrin Seemingly the thing circled on the left here (i.imgur.com/AArZwya.png). But my question is about how to apply this so it's a bit fuzzy to me.
the net matrix is upper triangular because our first basis element is an eigenvector giving a single entry in the $1$st position of column $1$. Then we took the rest of the basis to be the orthogonal complement to the eigenvector. We know there exists a basis $B^*$ which ensures the $n-1 \times n-1$ sub block is upper triangular. Since this sub-block acts only on orthonormal vectors to V, it creates an orthonormal basis which triangularises the sub-block.
OK, right. Note that if you were doing this argument for the spectral theorem (to get diagonalizability), you'd want to argue that the top-right block is $0$.
@TedShifrin Hmm, that actually does seem to check out if I just integrate $Q(t)$ from $0$ to $t$. How did you figure that out, as in, if I have a DE with an arbitrary number of variables and of arbitrary order, how do I know what integral to solve?
shin: for law school itself it's mostly just reading, with memorization happening if at all by osmosis. for state bar exams there's quite a lot of memorization. a number of companies sell state-specific review courses. big $$ tag on these, but my employer paid for mine.
ted some of the berries we see around here are the same distributor that you see in 'real' grocery stores, but maybe not what you see at a whole foods. the whole foods is a little further down the road and parking is impossible. i'll eat pesticide and feed it to my daughter to avoid an annoying parking situation
Whole Foods is definitely better, and Sprouts, too. I have been schlepping 25 miles each way to the farmers market on Sundays for berries and other things, but Sprouts is still OK.
The US is accelerating back to the 19th century. Rights will be disappearing like beer. Legal profession probably irrelevant because the autocratic thugs are in charge.
@TedShifrin Yes it does >:{ I'm not angry with you for saying it; That is indeed the direction. I'm angry with the idiots wanting to roll back time....
@leslietownes She's going to be a good radical, (i.e., speaking the truth, which now days is considered radical) like Greta Thunberg ... She's got your Berkeley genes, after all!
