pr.probability – Change variables in Gaussian integral over subspace $S$

I have been thinking about a problem and I have an intuition about it but I don’t seem to know how to properly address it mathematically, so I’m sharing it with you hoping to get help. Suppose I have two $ntimes n$ real matrices $C$ and $M$ and consider the Gaussian integral:
$$I = Nint e^{-frac{1}{2}ilangle x, C^{-1} xrangle} e^{langle x, M xrangle}dx$$
where $N$ is a normalizig constant and I’m writting:
$$langle x, A x rangle = sum_{i,j}x_{i}A_{ij}x_{j}$$
the inner product of $x$ and $Ax$ on $mathbb{R}^{n}$. $C$ is the covariance of the Gaussian measure; moreover, suppose $M$ is not invertible and has $1 le k < n$ linearly independent eigenvectors associated to the eigenvalue $lambda = 0$. All other eigenvectors of $M$ are also linearly independent, but associated to different nonzero eigenvalues.

This is my problem. I’d like to know how does the formula for the Gaussian integral $I$ changes if I was to integrate over the subspace $S$ spanned by the eigenvectors $v_{1},…,v_{k}$ associated to $lambda = 0$. Intuitively, this integral wouldn’t have the $e^{langle x, Mx rangle}$ factor because $Mequiv 0$ in this subspace. In addition, since $S$ is a $k$-dimensional subspace, I’d expect this integral would become some sort of Gaussian integral over a new variable $y$ which has now $k$ entries.

I would like to know if my intuition is correct and if it is possible to explicitly write this new integral over $S$, which I was not able to do by myself. Thanks!

hyperbolic geometry – Reference request: Discrete subgroup of $mathrm{PO}(n,1)$ preserving proper subspace has infinite covolume

I’m looking for a reference for the following claim:

$newcommand{PP}{mathrm{PO}(n,1)}$
Let $PP$ denote the group of isometries of $V = mathbb{R}^{n,1}$ preserving the upper sheet of the hyperboloid (i.e. $PP$ is the group of isometries of hyperbolic $n$-space), and $G$ a discrete subgroup of $PP$.

Claim. If $G$ preserves a proper (linear) subspace of $V$, then $G$ is of infinite covolume in $PP$.

The way I think a proof goes is:

  • If $G$ preserves a subspace $U$, it preserves its orthogonal complement $U^perp$ too.
  • Either $U$ or $U^perp$ either intersects the hyperboloid or contains an isotropic vector.
  • Up to swapping $U$ and $U^perp$, if $U$ intersects the hyperboloid, it preserves a proper subspace of hyperbolic space, and is therefore of infinite covolume.
  • Similarly, if $U$ only contains an isotropic vector, it fixes an ideal point of hyperbolic space, and must also be of infinite volume.

I’m not 100% convinced by this argument, and I looked in Ratcliffe’s and Vinberg’s book for a better one and/or a reference, to no avail.

Thanks!

fa.functional analysis – If $left(a otimes 1_B right) v$ span $mathcal{H}_A otimes mathcal{H}_B$, is the subspace $V ni v$ a product $V = V_A otimes mathcal{H}_B$?

Consider a subspace $V$ of $mathcal{H}_A otimes mathcal{H}_B$, with $mathcal{H}_A$ and $mathcal{H}_B$ finite-dimensional Hilbert spaces, such that $left( a otimes 1_B right) v$, with $v in V$, $a in mathcal{L} left( mathcal{H}_A right)$ span $mathcal{H}_A otimes mathcal{H}_B$. Can $V$ be written as $V_A otimes mathcal{H}_B$ for some subspace $V_A$ of $mathcal{H}_A$?

Thanks for the help with my related question.

linear algebra – Finding a basis given a set that spans a subspace

I have to solve this:

Let $W$ denote the subspace of $R^5$ consisting of all the vectors having coordinates that sum to zero. The vectors
$$u_1=(2,-3,4,-5,2), quad u_2=(-6,9,-12,15,-6), $$
$$u_3=(3,-2,7,-9,1), quad u_4=(2,-8,2,-2,6), $$
$$u_5=(-1,1,2,1,-3), quad u_6=(0,-3,-18,9,12), $$
$$u_7=(1,0,-2,3,-2), quad u_8=(2,-1,1,-9,7) $$
generate $W$. Find a subset of ${u_1,u_2,….,u_8}$ that is a basis for $W$

What I have done is try to find which of the vectors are linear combinations of others. I put the vectors on a matrix and found that four rows became filled with zeroes thus the basis would have only four elements i.e., $dim(W=4)$ and I don’t understand why?

My second question is if I can be sure that the vectors that did not become filled with zeroes are lineraly independent.

linear algebra – Height Of A Vector Respect To A SubSpace

Can anyone explain me what means the following:

To compute the volume of a $n$ dimensional parallelepiped we use the base x height rule, we pick a vector, then height is the distance from this vector to the subspace spanned by the remainimg vectors, where base is the $(n-1)$ dimensional volume formed by the remaining vectors.

¿What means height there?, I understand it with $2$, $3$ dimensions, but what about 4?

vector spaces – Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Is my answer ok? (Linear Algebra Done Right)

I am reading “Linear Algebra Done Right 3rd Edition” by Sheldon Axler.

This book contains the following exercise:

A function $f:mathbb{R}tomathbb{R}$ is called periodic if there exists a positive number $p$ such that $f(x) = f(x+p)$ for all $xinmathbb{R}$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.

I solved this exercise, but I am not sure that my answer is correct because my answer is too simple.
Is the following answer of mine right or not?

Define the function $f:mathbb{R}tomathbb{R}$ by $f(x) = 0$ for $xinmathbb{R}setminusmathbb{Z}$ and $f(x) = 1$ for $xinmathbb{Z}$.
Define the function $g:mathbb{R}tomathbb{R}$ by $g(x) = 0$ for $xinmathbb{R}setminussqrt{2},mathbb{Z}$ and $g(x) = 1$ for $xinsqrt{2},mathbb{Z}$.
$f$ is a periodic function whose period is $1$.
$g$ is a periodic function whose period is $sqrt{2}$.
Since $0inmathbb{Z}capsqrt{2},mathbb{Z}$, $(f+g)(0)=f(0)+g(0)=1+1=2$.
Let $p$ be any positive real number.
Since $pnotinmathbb{Z}capsqrt{2},mathbb{Z}$, so $(f+g)(0+p)=(f+g)(p)=f(p)+g(p) < 2$.
So, $(f+g)(0)neq (f+g)(0+p)$ for any positive real number $p$.
So, the function $f+g$ is not periodic.

general topology – The annihilator $M^perp$ of a set $M neq emptyset$ in an inner product space X is a closed subspace of X.

I’m trying to prove the following:

Show that the annihilator $M^perp$ of a set $M neq emptyset$ in an inner product space X is a closed subspace of X.

Next is the proof I have done, which indeed only proves that $M^perp$ can’t be open, from what I know, this doesn’t imply it is closed, so I need to ensure $M^perp$ is closed. I would really appreciate any corrections to this argument and if possible any orientation on how to prove the statement with basic concepts of the topic. Thanks.

Proof:

i) By definition
$$
M^perp = {x in X: langle x,y rangle = 0 quad forall y in M}
$$

Suppose $x_1,x_2~in M^perp$ and $alpha$ is a scalar. Then
$$
langle alpha x_1+x_2, y rangle = alpha langle x_1,y rangle +langle x_2, yrangle = 0
$$

This means $~alpha x_1 +x_2 ~in M^perp$, then $~M^perp$ is a subspace of X.

ii) Note that if $M={vec{0}}$, then $M^perp = X$ which is close and open, then in particular $M^perp$ is closed. Suppose then that M contains at least one element $yneq vec{0}$. On the other hand if $M^perp = {vec{0}}$, $M^perp$ is a singleton and therefore is closed. Suppose then that $M^perp$ contains at least one $xneq vec{0}$.

If $M^perp$ is an open subset of X and $x_0 in M^perp$ and $x_0neq vec{0}$, there exist a real number $epsilon>0$ such that the ball with the center in $x_0$ and radius $epsilon$ is contained in $M^perp$

$$
B(x_0, epsilon)={xin X: ||x-x_0||<epsilon} subset M^perp
$$

Now, consider the vector $x’ = x_0+beta y~$ where $yin M$, $y neq vec{0}$ and $beta$ is a scalar, it is true that we can choose $beta$ such that
$$
||x’-x_0||= ||x_0+beta y-x_0||=||beta y||=|beta|~||y||<epsilon longrightarrow x’ in B(x_0,epsilon)
$$

Since $yin M rightarrow langle x_0, yrangle = 0$, and therefore
$$
langle x’,yrangle = langle x_0+beta y, yrangle = langle x_0, yrangle + betalangle y,y rangle = beta langle y,yrangle neq vec{0}. ~Since ~y neq vec{0}.
$$

then $x’ notin M^perp$. In summary, we can always form $x’$ such that $x’ in B(x_0,epsilon)$ but $x’ notin M^perp$. This mean, $B(x_0,epsilon) notsubset M^perp$, which implies $x_0$ is not an interior point of $M^perp$, in consequence $M^perp$ is not open.

Linear algebra: Are vectors that define a game object considered to be a subspace?

In Linear-algebra a subspace is a collection of vectors that when added with each other or scaling would result in another vector in that same subspace.

How does subspaces relate to objects/players within the game world?

An object (player or enemy) in a game itself is just a collection of vertices that can be defined as a collection of vectors right? Does that mean that each object is also a subspace?

If I scale one object to be large enough to overlap/collide with another object does that mean those two objects are still in their own subspaces?

sg.symplectic geometry – Every half-dimensional subspace of a symplectic vector space has a Lagrangian complement

Let $(V, omega)$ be a finite-dimensional real symplectic vector space, i.e. $omega : V times V to mathbb{R}$ is a non-degenerate skew-symmetric bilinear map.
A linear subspace $L subset V$ is called Lagrangian if $L = L^perp$, where $L^perp = {v in V : omega(v, L) = {0}}$.

Let $U subset V$ be a linear subspace such that
$$dim U = frac{1}{2} dim V.$$
Question. Why is there always a Lagrangian subspace $L subset V$ such that $V = U oplus L$?

This is easy if $U$ is also Lagrangian, and there are plenty of sources explaining this (we can set $L = I(U)$ where $I$ is a compatible complex structure). But I didn’t find any reference explaining this more general fact, although I have seen it used in some research papers. It is used, for instance, for the existence of local Lagrangian bisections in symplectic groupoids. Any hints on the proof would be appreciated.