## graphs – Power of adjacency matrix

Let $$G$$ be a weighted graph with a weight function $$w longrightarrow mathbb{R}^{+}$$. Let $$G’$$ denotes the weighted matrix with adjacency matrix

$$A_{G’} = sum_{i=0}^{k} (xA)^{i}$$

where $$k$$ is integer and $$x$$ is a variable.

I am not getting what is $$A_{G’}$$ matrix? Is it contains all walks of length $$k$$ or is it something else?

## How to get non-zero entries of a sparse 3D matrix?

I have a sparse 3D matrix and I want to get the non-zero values. Is there any way to project it into some other space, where it would be easy to get those non-zero entries? Any answers or some source to read from is highly appreciated.

## plotting – I want to solve a differential equation in matrix form

I tried something like that. Z is my hamiltonian

``````n = 5;

σ2 = 0.1;

RR = RandomReal[{-Sqrt[3*σ2], Sqrt[3*σ2]}, n];

Z = Table[
KroneckerDelta[i - j + 1] + KroneckerDelta[i - j - 1], {i, 1,
n}, {j, 1, n}] + DiagonalMatrix[RR];

usol = NDSolveValue[{I D[ψ[x, t], t] ==
Z.ψ[x, t], ψ[0, t] == 0, ψ[n, t] == 0}, ψ, {t,
0, 1}]
``````

I solve the problem finding the eigenstates and eigenvalues, then I choose a one site and evolve in time but now I want to solve the differential equation directly to check my results.

## linear algebra – A problem about determinant and matrix

Suppose $$a_{0},a_{1},a_{2}inmathbb{Q}$$, such that the following determinant is zero, i.e.

$$left |begin{array}{cccc}\ a_{0} &a_{1} & a_{2} \ \ a_{2} &a_{0}+a_{1} & a_{1}+a_{2} \ \ a_{1} &a_{2} & a_{0}+a_{1}\ end{array}right| =0$$

Show that $$a_{0}=a_{1}=a_{2}=0$$

I think it’s equivalent to show that the rank of the matrix is 0, and it’s easy to show the rank cannot be 1.

But I have no idea how to show that the case of rank 2 is impossible. So is there any better idea? Thanks.

## linear algebra – Matrix for rotation around a vector without using rodrigues rotation formula

for example how can I write the rotation matrix around the vector $$vec v = (1,1,1)$$ with the angle 90°.

I searched all the other questions and they all say that rodrigues rotation formula is the way to go but I’m wondering if there’s an easy way for me.

thanks.

## linear algebra – How can I find the inverse of a matrix with undetermined variables?

all. I am going to find the inverse of the matrix after the derivatives. But, the system terminates the calculation during the process. So, what is the problem inside? How can I fix it? Many thanks!

The code can be found from the following link.

## matrix – Invert parent transform (doesn’t work for combination of rotation and scale)

### My problem

I’m working with Qt3D and my problem is almost exactly like this one:

https://stackoverflow.com/q/60995155/3405291

### Suggested solution

A solution is suggested here:

https://stackoverflow.com/a/61315454/3405291

### Understanding the solution

I have a problem understanding the suggested solution. Specifically:

The problem is that the `QTransform` node does not store the transformation as a general `4x4` matrix. Rather is decomposes the matrix into a `3` transformations that are applied in fixed order:

S – a diagonal scaling matrix

R – the rotation matrix

T – translation

and then applies it in the order `T * R * S * X` to a point `X`.

So when the transformation on the parent is `M = T * R * S`, then the inverse on the child will be `M^-1 = S^-1 * R^-1 * T^-1`. Setting the inverse on the child `QTransform` will attempt to decompose it in the same way:

`M^-1 = T_i * R_i * S_i = S^-1 * R^-1 * T^-1`

That doesn’t work, because particularly `S` and `R` don’t commute like this.

I don’t understand the above assertions. Can anyone explain them to me. Just help me realize.

## matrices – Applying repeated doubling to an update Matrix

Given a rule to obtain a new $$x,y$$ positions from an initial $$x_0,y_0$$

The rule is
$$binom{x’}{y’} = begin{pmatrix} v_x & -v_y \ v_y & v_x end{pmatrix} binom{x}{y}$$

$$x’ = v_x x – v_y y$$

$$y’ = v_y x + v_x y$$

$$f(x,y) = (v_x x – v_y y, v_y x + v_x y)$$

Now let’s say I wanted to use this rule to update a position $$x_0,y_0$$

How can I use repeated doubling to calculate the position after updating it $$n$$ times without having to manually update the position $$n$$ times

For instance lets say we started at position $$(-10, 0)$$ and we wanted to update this position 5 times and our given value for $$(u_x,u_y) = (frac{1}{2},frac{1}{2})$$

after the first update(n=1) the position would be $$(-5,-5)$$ and then the 5th update(n=5) the position would be $$( frac{5}{4}, frac{5}{4})$$

How could I calculate the position for $$( frac{5}{4}, frac{5}{4})$$ without having to calculate n = 2 or n= 3 or n= 4 or simply just using the concept of repeated doubling

## linear algebra – Eigenvalues of scaling of matrix

Let $$A$$ be a (real or complex) square matrix, let $$alpha neq 0$$.

Is it true that $$lambda$$ is an eigenvalue of $$A$$ if and only if and only if $$alpha lambda$$ is an eigenvalue of $$alpha A$$?

I think yes, here is why I suppose so: $$lambda$$ is an eigenvalue of $$A ifflambda I- A$$ is non injective $$iff alpha lambda I – alpha A$$ is non injective $$iff alpha lambda$$ is an eigenvalue of $$alpha A$$.

Is this correct?

## pr.probability – Distribution and Expectation of Inverse of a Random Bernoulli Matrix

This question cropped up as a part of my research. Let us assume a $$ntimes n$$ random matrix $$mathbf{M}$$ with elements iid distributed to a Bernoulli distribution that takes values $${0,1}$$ with probability $$p = 1/2$$.

What I want to know is that what sort of distribution would $$mathbf{M}^{-1}$$ have? and what could its possible expectation be?

My guess so far has been that the distribution remains the same. I have posted this question on math.stackexchange too.