Simplifying Linear Algebra: Techniques, Solutions, and Examples

Vector projection in polar cordinates
In Euclidean Space and Cartesian coordinates system, we know the vector projection of Vector u onto a vector v is simply
$\stackrel{\to <!--\; \to \; -->}{P}=\frac{(\stackrel{\to <!--\; \to \; -->}{u}.\stackrel{\to <!--\; \to \; -->}{v})\stackrel{\to <!--\; \to \; -->}{v}}{\stackrel{\to <!--\; \to \; -->}{v}.\stackrel{\to <!--\; \to \; -->}{v}}$
What would be the vector projection on polar or spherical coordinates or alternate coordinate systems?
Suppose we have vectors defined as
$\stackrel{\to <!--\; \to \; -->}{u}=u^{\alpha }{e}_{\alpha }$
$\stackrel{\to <!--\; \to \; -->}{v}=v^{\beta }{e}_{\beta }$
Does a projection vector of u on v becomes
$\stackrel{\to <!--\; \to \; -->}{P}=\frac{(u^{\alpha }{v}^{\beta }{e}_{\alpha \beta })\stackrel{\to <!--\; \to \; -->}{v}}{v^{\alpha }{v}^{\beta }{e}_{\alpha \beta }}$
Where $e_{\alpha }$ represent the basis vector and $e_{\alpha \beta }$ the metric tensor.

Polar Coordinate Transformation - Motivation
I am trying to work out the reason why the integral
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->(x^2+y^2)}dxdy$
is, in polar coordinates,
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->r^2}rdrd\mathrm{\theta <!--\; \theta \; -->}$
As I understand it, a polar coordinate transformation involves the following substitution:
$(x,y)\to <!--\; \to \; -->(r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},r\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$
This would imply that
$-<!--\; -\; -->(x^2+y^2)=-<!--\; -\; -->((r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2+(r\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2)=-<!--\; -\; -->r^2((\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2+(\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2)=-<!--\; -\; -->r^2$
which gets us this far
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->r^2}dxdy$
To motivate why
$dxdy=rdrd\mathrm{\theta <!--\; \theta \; -->}$
I thought of the following argument:
$$\begin{gathered} dx=d(x(\mathrm{\theta <!--\; \theta \; -->},r))=\frac{\mathrm{\partial <!--\; \partial \; -->}(x(\mathrm{\theta <!--\; \theta \; -->},r))}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\theta <!--\; \theta \; -->}}d\mathrm{\theta <!--\; \theta \; -->}+\frac{\mathrm{\partial <!--\; \partial \; -->}(x(\mathrm{\theta <!--\; \theta \; -->},r))}{\mathrm{\partial <!--\; \partial \; -->}r}dr+\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2(x(\mathrm{\theta <!--\; \theta \; -->},r))}{(\mathrm{\partial <!--\; \partial \; -->}r{)}^2}(dr{)}^2+... \\ =r(-<!--\; -\; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}+dr\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+... \end{gathered}$$
$dy=r(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}+dr\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+...$
$$\begin{gathered} \therefore <!--\; \therefore \; -->dxdy=[r(-<!--\; -\; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}+dr\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+...][r(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}+dr\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+...] \\ =rdrd\mathrm{\theta <!--\; \theta \; -->}((\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2-<!--\; -\; -->(\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2)+... \\ =rdrd\mathrm{\theta <!--\; \theta \; -->}((\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2-<!--\; -\; -->(\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2),\text{ ignoring }\mathit{\text{o}}((dr{)}^2)\text{ and }\mathit{\text{o}}((d\mathrm{\theta <!--\; \theta \; -->}{)}^2)\text{ terms.} \end{gathered}$$
However, I am off by a minus sign, which would enable me to argue
$dxdy=rdrd\mathrm{\theta <!--\; \theta \; -->}((\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2+(\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2)=rdrd\mathrm{\theta <!--\; \theta \; -->}$
If it were correct, I would find this line of argument would be much more analytically convincing than the typical argument involving
$dxdy=dA=rdrd\mathrm{\theta <!--\; \theta \; -->}$
which I find to be less mechanistically obvious than the above substitution-based argument that I considered but that I am not being able to fully justify.
Could you please tell me whether my substitution-based argument can work, potentially by correcting some mistake or another that I might have made? If not, do you have any similarly analytical or mechanistic justification as to why $dxdy=rdrd\mathrm{\theta <!--\; \theta \; -->}$ ?

What is the Mathematical equivalent of the definition of vectors in physics?
Physics has an additional requirement to define physical quantities as vectors i.e. if they are invariant under rotation of the coordinate system for example quantities like force, velocity etc.
How does it correspond to the mathematical formulation? Is it a subset of vector spaces? Is it an alternate characterisation to the axioms of vector space and can it be proved from the former axioms?
From wikipedia:
A vector space over a field F is a set V together with two operations that satisfy the eight axioms.
The first operation, called vector addition or simply addition + : V × V$\to <!--\; \to \; -->$ V, takes any two vectors v and w and assigns to them a third vector which is commonly written as v + w, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set V ). The second operation, called scalar multiplication · : F × V$\to <!--\; \to \; -->$ V， takes any scalar a and any vector v and gives another vector av. (Similarly, the vector av is an element of the set V ).
Please try to respond in layman's terms as I am just starting with university level maths. If the proof involves advanced concepts, I am satisfied with reference to the topic in which I might encounter this in advanced studies.

Deriving the distance of closest approach between ellipsoid and line (prev. "equation of a 3-dimensional line in spherical coordinates")
Currently trying to solve a problem of calculating the smallest distance between a given ellipsoid centered on the coordinate system starting point and a given line (located... somewhere).
After chasing a few promising but non-functional methods I have settled on trying to use spherical coordintates - to determine the formula for said distance by determining the formula of the distance of the ellipsoid from the center and doing the same for the line, consequently subtracting the two, and using gradient descent on the resulting function to approach a minimum (hopefully the global one).
However, while that makes the ellipsoid calculations easy, I have found no consise way of determining a line in spherical coordinates in equation form. I have found the old questions on similar topics proposing use of Euler angles and the like, but that does not seem to be the solution (possibly because I haven't managed to appreciate it). So, asking here - is there any way to derive an equation for a line in 3-dimensional space in spherical coordinates?
Alternate methods for the task at hand are appreciated too - for the record, my previous lead was using a cyllindrical coordinate system with the line as its X axis, but the resulting formula for the ellipsoid turned out to be bogus.
Edit: may have figured out a solution for the bigger problem that does not rely on the spherical equation - see my own answer to the question. Title changed accordingly.
Edit 2: Scratch that. Fell into the same trap again; that solution is not going to work.

Using index notation to write $d^2=0$ in terms of a torsion free connection.
Let (M,g) be a Riemannian manifold and let $\mathrm{\omega <!--\; \omega \; -->}$ be a 1-form on M. I want to rewrite $d^2\mathrm{\omega <!--\; \omega \; -->}=0$ in terms of the Levi-Civita connection.
I can show the following:
$d\mathrm{\omega <!--\; \omega \; -->}(X,Y)=({\mathrm{\nabla <!--\; \nabla \; -->}}_X\mathrm{\omega <!--\; \omega \; -->})(Y)-<!--\; -\; -->({\mathrm{\nabla <!--\; \nabla \; -->}}_Y\mathrm{\omega <!--\; \omega \; -->})(X),$
which in index notation reads
$(d\mathrm{\omega <!--\; \omega \; -->}{)}_{ab}=2{\mathrm{\nabla <!--\; \nabla \; -->}}_{[a}{\mathrm{\omega <!--\; \omega \; -->}}_{b]}.$
Similiarly for a 2-form, $\mathrm{\mu <!--\; \mu \; -->}$, we have:
$d\mathrm{\mu <!--\; \mu \; -->}(X,Y,Z)=({\mathrm{\nabla <!--\; \nabla \; -->}}_X\mathrm{\mu <!--\; \mu \; -->})(Y,Z)-<!--\; -\; -->({\mathrm{\nabla <!--\; \nabla \; -->}}_Y\mathrm{\mu <!--\; \mu \; -->})(X,Z)+({\mathrm{\nabla <!--\; \nabla \; -->}}_Z\mathrm{\mu <!--\; \mu \; -->})(X,Y),$
which in index notation reads
$(d\mathrm{\mu <!--\; \mu \; -->}{)}_{abc}=3{\mathrm{\nabla <!--\; \nabla \; -->}}_{[a}{\mathrm{\varphi <!--\; \varphi \; -->}}_{bc]}.$
Now plugging in $d\mathrm{\omega <!--\; \omega \; -->}$ for $\mathrm{\mu <!--\; \mu \; -->}$ we get
$0=d^2\mathrm{\omega <!--\; \omega \; -->}=(d(d\mathrm{\omega <!--\; \omega \; -->}){)}_{abc}={\mathrm{\nabla <!--\; \nabla \; -->}}_{[a}(d\mathrm{\omega <!--\; \omega \; -->}{)}_{bc]}.$
I want to plug in the above expression (in index notation) for dω but I'm not really sure how to handle the indices. Do I just get
$3{\mathrm{\nabla <!--\; \nabla \; -->}}_{[a}2{\mathrm{\nabla <!--\; \nabla \; -->}}_{[b}{\mathrm{\omega <!--\; \omega \; -->}}_{c]]}=6{\mathrm{\nabla <!--\; \nabla \; -->}}_{[a}{\mathrm{\nabla <!--\; \nabla \; -->}}_b{\mathrm{\omega <!--\; \omega \; -->}}_{c]}?$

Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?
A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

The Determinant, Tensors, and Orientation
I am a bit confused about orientation and tensors as exemplified by the determinant.
If we have an inner product or a metric and transform a vector from one coordinate system to another the magnitude of that vector is unchanged as the coordinate representation of the metric also changes. Therefore it is natural to think of a vector as a tensor and it's length doesn't change under orientation reversing transformations.
In contrast, we can consider the determinant as an alternating tensor from the exterior algebra, for instance in ${\mathbb{R}}^3$,
$det(v_1,v_2,v_3)=e^1\wedge <!--\; \wedge \; -->e^2\wedge <!--\; \wedge \; -->e^3(v_1,v_2,v_3)$
But if we perform a reflection, A then we get,
$det(A{v}_1,A{v}_2,A{v}_3)=det(A)det(v_1,v_2,v_3)=-<!--\; -\; -->det(v_1,v_2,v_3)$
Does that mean that the 3-covector $e^1\wedge <!--\; \wedge \; -->e^2\wedge <!--\; \wedge \; -->e^3$ when contracted with 3 vectors is not invariant, which would seem to violate the idea of contracting tensors to a scalar produces an invariant under transformations.
Or do we instead by transforming our 3-covector and the input vectors get an invariant oriented volume measure?
$e^{\prime 1}\wedge <!--\; \wedge \; -->e^{\prime 2}\wedge <!--\; \wedge \; -->e^{\prime 3}(A{v}_1,A{v}_2,A{v}_3)=e^1\wedge <!--\; \wedge \; -->e^2\wedge <!--\; \wedge \; -->e^3(v_1,v_2,v_3)?$
That would seem to mean that the determinant includes an arbitrary sign convention and that the definition of the determinant is different in left and right handed oriented coordinate systems.

How to prove an expression to be a tensor?
How to prove that the expression ${\mathrm{\phi <!--\; \phi \; -->}}_{,ij}:=\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\phi <!--\; \phi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}_i\mathrm{\partial <!--\; \partial \; -->}{x}_j}=\mathrm{\nabla <!--\; \nabla \; -->}\mathrm{\nabla <!--\; \nabla \; -->}\mathrm{\phi <!--\; \phi \; -->}$ is a tensor of second order where φ is a scalar? Furthermore, how to prove that $q_{j_1\cdots <!--\; \cdots \; -->j_n}$ is a vector?
We can either prove it by definition or use the so-called "tensor recognition theorem" claiming that if $q_{j_1\cdots <!--\; \cdots \; -->j_n}$, then p must be a tensor of order m+n, where $q_{j_1\cdots <!--\; \cdots \; -->j_n}$ is a tensor of order n and $r_{i_1\cdots <!--\; \cdots \; -->i_m}$ a tensor of order m.

Can vectors even be expressed unambiguously?
Vectors are abstract concepts. Lets take one of the simplest, more concrete vectors out there: an euclidian vector in 2D. Now, I think that even such a vector is abstract, it cannot be written down, it cannot be expressed, it cannot be specified, it cannot be conveyed. At best you can give it a name, like V.
What one could try to do, is to express it as a linear combination of other vectors of that space, for example an orthonormal basis $B_2\}$.
For example, it may be that "$V=28\cdot <!--\; \cdot \; -->B_1+7\cdot <!--\; \cdot \; -->B_2$" Or, "V=(28,7) in the basis $\{B_1,B_2\}$"
The problem is that I expressed the vector in terms of other vectors. (28,7) means nothing unless I can somehow describe $B_2$ and $B_2$. After all, if I chose another basis $\{C_1,C_2$, (28,7) would represent a completely different vector.
And I can't describe $B_1$ or $B_2$, express them, other than by doing so in terms of other vectors, just like I couldn't do it for V.
So I cannot specify which vector V is, other than adding two new vectors, which I also can't specify. It all seems completely circular to me.
Trying to frame this as a question: how can one write down a vector in a way that it actually specifies which vector it is? How can someone even specify what basis he is using? Aren't all those expressions circular and meaningless?

Find the Coordinates of the foot of the perpendicular from the point A with Coordinates ( 3,5 ) to the line -5x+2y-5=0 .
1. What is the x coordinate of the foot of the perpendicular? Giveanswer as a number only correct to at least 3 decimal places.
2. What is the y coordinate of the foot of the perpendicular? Giveanswer as a number only correct to at least 3 decimal places.
3. Find the distance from the point A to the foot of theperpendicular. Give answer as a number only correct to at least 3 decimal places

Formula relating covariant derivative and exterior derivative
According to Gallot-Hulin-Lafontaine one has
$d\mathrm{\alpha <!--\; \alpha \; -->}(X_0,\cdots <!--\; \cdots \; -->,X_q)=\underset{i=0}{\overset{q}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^i{D}_{X_i}\mathrm{\alpha <!--\; \alpha \; -->}(X_1,\cdots <!--\; \cdots \; -->,X_{i-<!--\; -\; -->1},X_0,X_{i+1},\cdots <!--\; \cdots \; -->,X_q)$
It seems to me that it should be
$d\mathrm{\alpha <!--\; \alpha \; -->}(X_0,\cdots <!--\; \cdots \; -->,X_q)=\underset{i=0}{\overset{q}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^i{D}_{X_i}\mathrm{\alpha <!--\; \alpha \; -->}(X_0,\cdots <!--\; \cdots \; -->,\stackrel{^<!--\; ^\; -->}{X_i},\cdots <!--\; \cdots \; -->,X_q)$
Is this right ?

Equivalence of statements of orientation of a manifold
a) $\mathrm{\exists <!--\; \exists \; -->}$ a nowhere vanishing top form $\mathrm{\alpha <!--\; \alpha \; -->}$ on M
b)There exist a system of coordinate neighborhoods covering M and such that the Jacobian matrix corresponding to the change of corrdinates where two patches overlap, has positive determinant on every overlap.
c) The space of alternating top forms ${\mathrm{\Lambda <!--\; \Lambda \; -->}}^N{T}^{\ast <!--\; \ast \; -->}M$, is isomorphic to $M\times <!--\; \times \; -->\mathbb{R}$
I am struggling to convince myself that $a)⟹<!--\; ⟹\; -->b)$, and I think I am missing something really obvious.
If $\mathrm{\alpha <!--\; \alpha \; -->}=a(x_i)d{x}_1\wedge <!--\; \wedge \; -->...\wedge <!--\; \wedge \; -->d{x}_n$ in one set of coordinates, then under a coordinate transformation
$\mathrm{\alpha <!--\; \alpha \; -->}=a(x_i)det(\frac{\mathrm{\partial <!--\; \partial \; -->}{x}_i}{\mathrm{\partial <!--\; \partial \; -->}{x}_j^{\prime }})d{x}_1^{\prime }\wedge <!--\; \wedge \; -->...\wedge <!--\; \wedge \; -->d{x}_n^{\prime }$
But if the determinant changes sign for at least one overlap of coordinate neighborhoods for every manifold covering , I don't see why this means the top form must vanish. Since we put the charts onto the manifold, why can the "observer" not merely say that their choice of coordinates is such that the one form is positive in one choice, and negative in another?

Geometric intuition about the exterior derivative
Let M be a smooth manifold. One k-form is a section of the bundle $\stackrel{k}{\bigwedge <!--\; \bigwedge \; -->}{T}^{\ast <!--\; \ast \; -->}M$, that is, if $p\in <!--\; \in \; -->M$ and $\mathrm{\omega <!--\; \omega \; -->}$ is a k-form then ${\mathrm{\omega <!--\; \omega \; -->}}_p$ is one k-linear alternating real function defined in the cartesian product of k copies of the tangent space at p.
Given ω one can then build a new differential k+1 form called the exterior derivative $d\mathrm{\omega <!--\; \omega \; -->}$. This can be easily defined in a coordinate system (x,U) as
$d\mathrm{\omega <!--\; \omega \; -->}=\underset{}{\sum <!--\; \sum \; -->}d{\mathrm{\omega <!--\; \omega \; -->}}_{i_1\dots <!--\; \dots \; -->i_k}\wedge <!--\; \wedge \; -->d{x}^{i_1}\wedge <!--\; \wedge \; -->\cdots <!--\; \cdots \; -->\wedge <!--\; \wedge \; -->d{x}^{i_k}$
where ${\mathrm{\omega <!--\; \omega \; -->}}_{i_1\dots <!--\; \dots \; -->i_k}$ are the components of ω in the coordinate system (x,U). One then can easily show this definition doesn't depend on the coordinate system chosen. Based on that it is possible to show many properties of the exterior derivative then.
Now, the problem is that I simply can't get one geometrical intuition on what the exterior derivative is. As I know, a k-form can be thought of as an object which performs measurements on k-vectors, that is, an object capable of measuring projections of k-vectors. This point of view comes from the fact that $\stackrel{k}{\bigwedge <!--\; \bigwedge \; -->}{V}^{\ast <!--\; \ast \; -->}$ is isomorphic to the space of linear functions on $\stackrel{k}{\bigwedge <!--\; \bigwedge \; -->}V$.
In that setting, what is the geometrical intuition behind the exterior derivative? Given one k-form how can one understand from one intuitive point of view what its exterior derivative is?

How to show that the Billiard flow is invariant with respect to the area form $\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})d\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->dt$
Consider a plane billiard table $D\subset <!--\; \subset \; -->{\mathbb{R}}^2$ (i.e. a bounded open connected set) with smooth boundary $\mathrm{\gamma <!--\; \gamma \; -->}$ being a closed curve. Next, let M denote the space of tangent unit vectors (x,v) with x on $\mathrm{\gamma <!--\; \gamma \; -->}$ and v being a unit vector pointing inwards. We then define the billiard map
$T:M\to <!--\; \to \; -->M.$
To understand the map T, we consider a point mass traveling from x in direction v. Let $x_1$ be the first point on $\mathrm{\gamma <!--\; \gamma \; -->}$ that this point mass intersects and suppose that v1 is the new direction of the mass upon incidence. Then T maps (x,v) to $(x_1,v_1)$.
We now introduce an alternate ''coordinate system'' describing M. Parametrize $\mathrm{\gamma <!--\; \gamma \; -->}$ by arc-length t and fix a point $(x,v)\in <!--\; \in \; -->M$. We can find t such that $x=\mathrm{\gamma <!--\; \gamma \; -->}(t)$ and let $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->(0,\mathrm{\pi <!--\; \pi \; -->})$ be the angle between the tangent line at x and v. The tuple $(t,\mathrm{\alpha <!--\; \alpha \; -->})$ uniquely determines the point (x,v) in M, and thus offers and alternative description of this space.
My question is as follows: I want to show that the area form given by
$\mathrm{\omega <!--\; \omega \; -->}:=\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->\mathrm{d}t$
is invariant under T.
I found a proof of this invariance property proof in S. Tabachnikov's Geometry and billiards but I'm having some trouble understanding a critical part of the proof.
If anyone can explain the proof to me (or provide me with another proof) I would highly appreciate it. An intuitive explanation is also appreciated, but I am looking for a rigorous proof if possible. We restate this theorem formally below and provide the proof as given by Tabachnikov.
Theorem 3.1. The area form $\omega =\sin <!--\; \; -->\alpha d\alpha \wedge <!--\; \wedge \; -->dt$ is T-invariant.
Proof. Define $f(t,t_1)$ to be the distance between $\mathrm{\gamma <!--\; \gamma \; -->}(t)$ and $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$. The partial derivative $\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{t}_1}$ is the projection of the gradient of the distance $\left|\mathrm{\gamma <!--\; \gamma \; -->}(t)\mathrm{\gamma <!--\; \gamma \; -->}(t_1)\right|$ on the curve at point $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$. This gradient is the unit vector from $\mathrm{\gamma <!--\; \gamma \; -->}(t)$ to $\mathrm{\gamma <!--\; \gamma \; -->}(t_1)$ and it makes angle ${\mathrm{\alpha <!--\; \alpha \; -->}}_1$ with the curve; hence $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}{t}_1=\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1$. Likewise, $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}t=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}$. Therefore,
$\mathrm{d}f=\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{d}t+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{t}_1}\mathrm{d}{t}_1=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}t+\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1\mathrm{d}{t}_1$
and hence
$0={\mathrm{d}}^{2}f=\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{d}\mathrm{\alpha <!--\; \alpha \; -->}\wedge <!--\; \wedge \; -->\mathrm{d}t-<!--\; -\; -->\sin <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1\mathrm{d}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\wedge <!--\; \wedge \; -->\mathrm{d}{t}_1.$
This means that $\mathrm{\omega <!--\; \omega \; -->}$ is a T-invariant form.
The above proof is copied directly from the book. I have the following questions about his method:
Is the domain of f the set $M\times <!--\; \times \; -->M$?
In the proof, are we specifically considering $(t,\mathrm{\alpha <!--\; \alpha \; -->})$ and $(t_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_1)$ such that $T(t,\mathrm{\alpha <!--\; \alpha \; -->})=(t_1,{\mathrm{\alpha <!--\; \alpha \; -->}}_1)$?
I am having a hard time understanding how the author obtains $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}{t}_1=\cos <!--\; \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_1$ and $\mathrm{\partial <!--\; \partial \; -->}f/\mathrm{\partial <!--\; \partial \; -->}t=-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}$. The explanation given feels mostly heuristic, how could I go about constructing a rigorous proof?

Showing that $detA=detB\cdot <!--\; \cdot \; -->detC$ when B,C are the restrictions of A onto a subspace
I am a bit unsure about one approach that is mentioned to prove this determinant result.
Here is the quote from Pages 100-101 of Finite-Dimensional Vector Spaces by Halmos:
Here is another useful fact about determinants. If $\mathcal{M}$ is a subspace invariant under A, if B is the transformation A considered on $\mathcal{M}$ only, and if C is the quotient transformation $A/\mathcal{M}$, then
$detA=detB\cdot <!--\; \cdot \; -->detC$
This multiplicative relation holds if, in particular, A is the direct sum of two transformations B and C. The proof can be based directly on the definition of determinants, or, alternatively, on the expansion obtained in the preceding paragraph.
What I am confused about is how you can use the definition of determinants to conclude this result.
In this book, the determinant of a linear transformation A is defined as the scalar $\mathrm{\delta <!--\; \delta \; -->}$ such that ${\mathrm{\alpha <!--\; \alpha \; -->}}_{ij}$ for all alternating n-linear forms w, where V is an n-dimensional vector space.
It is then shown that by fixing a coordinate system (or basis) and letting ${\mathrm{\alpha <!--\; \alpha \; -->}}_{ij}$ be the entries of the matrix of the linear transformation under the coordinate system, the determinant of the linear transformation A in that coordinate system is:
$detA=\underset{\mathrm{\pi <!--\; \pi \; -->}}{\sum <!--\; \sum \; -->}(\text{sgn}\mathrm{\pi <!--\; \pi \; -->}){\mathrm{\alpha <!--\; \alpha \; -->}}_{\mathrm{\pi <!--\; \pi \; -->}(1),1}\cdots <!--\; \cdots \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_{\mathrm{\pi <!--\; \pi \; -->}(n),n}$
where the summation goes over all permutations in ${\mathcal{S}}_n$.
I have been able to use the expression involving the coordinates to show this result, but I am not sure about how this would be done directly from the definition. I have tried looking at defining other alternating forms and using their product to show this, but I was not able to make much use of that approach.
Are there any suggestions for proving this result directly from the definition?
Edit: I would like to add that part of my confusion may be from the fact that A, B and C are all linear transformations on different vector spaces and I am not sure how the definition can be used in this situation.

Solving an Energy Minimization Problem, as Outlined in a Paper
The paper fairly succinctly boils the problem down to the following equation:
$\frac{1}{2}{x}^T(M+h^{2}L)x-<!--\; -\; -->h^2{x}^{T}Jd+x^{T}b$
Where:
- the system consists of m nodes and s springs
- h is the timestep(constant)
- x is a column vector of length 3m (or 3m∗1 matrix) representing the node positions,
- $(M+h^{2}L)$ is a matrix of size 3m∗3m (which, in this case, is symmetric, positive semi-definite, and does not change after being initialized.),
- J is a matrix of size 3m∗3s, and
- d is a column vector of size 3s representing the spring directions. (this particular equation is equation 14 in section 4 of the paper)
- b is a column vector of length 3m, which represents the xyz components of the external forces acting on each node
I am at a stage where I can generate all of vectors and matrices in code based on some arbitrary mesh consisting of nodes and links, but I am having trouble figuring out how I would go about solving for d and x. I have an initial guess for x (as outlined towards the ends of both section 3 and 4), but I don't have much experience solving entire systems like this.
My main question has two parts:
is there some generic procedure one would use to go about solving such a system?
if not, could some resources on the topic be suggested? Even some search terms more specific than "energy minimisation" would be of great use!
Ultimately I just don't know where to start, and the paper seems to simply say "starting with an initial guess for x, first we compute the optimal d", but I don't know what "compute the optimal d" entails.
Similarly, I wouldn't know where to start with computing the optimal values for x. I assume this would involve differentiating w.r.t. x and finding the values when the derivative equals zero but again, I have no idea how to apply this to entire matrices of values, as opposed to single values.
I am aware that each term in the above equation should evaluate to a scalar value, but I don't know how I would be able to obtain a value for each row/cell in the column vectors x and d.
NOTE: I believe there are one or two minor errors in the paper, such as $d\in R^{2s}$ which, I believe, should be $d\in R^{3s}$, since we are dealing with springs in 3 dimensions.

How do I find a rotation that brings this equation of a cone to standard axes?
I know for a vector in standard basis $\stackrel{¯<!--\; ¯\; -->}{x}$ we can find a representation for the same thing in a base B, denoted $[\stackrel{¯<!--\; ¯\; -->}{x}{]}_B$B by finding $\stackrel{¯<!--\; ¯\; -->}{C}[\stackrel{¯<!--\; ¯\; -->}{x}{]}_b=\stackrel{¯<!--\; ¯\; -->}{x}$. So far practice problems have been simple manipulations of matrices, so I'm not sure how to approach this problem:
A conic in the xy coordinate system is $5x^2-<!--\; -\; -->2\sqrt{3}xy+7y^2=16$. Find a rotation of coordinates that bring it to standard form. What is the matrix of this rotation? What is the cosine of the angle of the rotation?

How do you find the cartesian equation for ${\displaystyle r=\frac{3}{1+\cos T}}$

Find the inverse, if it exists, for the given matrix
$\left[\begin{array}{cc}3& 2\\ 2& 5\end{array}\right]$

Multivariable integral, probably related to the gamma function
Let $x={[x_1,x_2,..,x_n]}^T$ represent the vector of all variables and D be a diagonal matrix, the question is to integrate or give an approximate answer: $\int <!--\; \int \; -->\cdots <!--\; \cdots \; -->{\int <!--\; \int \; -->}_{[0,\mathrm{\infty <!--\; \infty \; -->}{]}^n}(x^{T}Dx{)}^{\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->1}exp(-<!--\; -\; -->x^{T}Dx)d{x}_1\dots <!--\; \dots \; -->d{x}_n$