Simplifying Linear Algebra: Techniques, Solutions, and Examples

I have a system of non-linear equations of the form
$A{x}_1^B\exp <!--\; \; -->(\frac{-<!--\; -\; -->C}{x_1})=k_1$
$A{x}_2^B\exp <!--\; \; -->(\frac{-<!--\; -\; -->C}{x_2})=k_2$
$A{x}_3^B\exp <!--\; \; -->(\frac{-<!--\; -\; -->C}{x_3})=k_3$
where $[x_1,x_2,x_3]$ and $[k_1,k_2,k_3]$ are known. The couple of constants $[A,B,C]$ is the unknown. The solution of this non-linear system of equations is given here: Solving a system of non linear equations
We must now ensure that $B<0$ at all times. How would you find one couple $[A^{\prime },B^{\prime },C^{\prime }]$ that best approach the solution of the system, with $B^{\prime }<0$.

Rotate a triangle with vertices (2, 1). (4. 1). (3, 3) by 90 degrees (clock wise) and then Translate it by Tx 4 and Ty 3 and finally rotate it by 30 degrees (anti-clock wise). (cos90=0, sin90 1. cos 30 = 0.866, sin30 =0.5)
T1 is Rotation with 90 T2 is Translation T3 is Rotation with 30
What is correct order of transformation matrix to multiply the metrix?
T3 T2 T1 or T1 T2 T3?
Should i follow the order given by the question or reverse order?

Representing a Linear Transformation as a Matrix
$T(\left[\begin{array}{ccc}a& & b\\ c& & d\end{array}\right])=\left[\begin{array}{ccc}d& & a\\ b& & c\end{array}\right]$
I attempted to apply $T$ to the standard basis of the $2\times <!--\; \times \; -->2$ matrices. However, I ended up with a $4\times <!--\; \times \; -->4$ matrix (understandable, as there are $4$ matrices in that basis). This matrix I came up with was
$\left[\begin{array}{ccccccc}0& & 0& & 0& & 1\\ 1& & 0& & 0& & 0\\ 0& & 1& & 0& & 0\\ 0& & 0& & 1& & 0\end{array}\right]$

representative matrix of a linear transformation
Let $T:{\mathbb{F}}^2\to <!--\; \to \; -->{\mathbb{F}}^2$ a linear tranformation such that $T^2=0$, Prove that $T=0$, or there exists a basis of $M_2(\mathbb{F})$ in which the representative matrix is: $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$.

I have a given $2\times <!--\; \times \; -->2$ special linear matrix, for example
$m=\left(\begin{array}{cc}55& 8469\\ 1& 154\end{array}\right)$
and I would like to get the generating form of it from the s and t matrices which are these:
$t=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)s=\left(\begin{array}{cc}0& -<!--\; -\; -->1\\ 1& 0\end{array}\right)$
I don't know if exist an algoritmh for this problem or any proceedings which could help me.

On the one hand, we always use the multiplication by transformation matrix when we want to receive a matrix in a new basis:
- $A^{\prime }=CA$ , where $A^{\prime }$ is a matrix in a new basis; $C$ is a transformation matrix; $A$ is a matrix in the standard basis.
However, when we speak about getting the matrix in its eigenbasis, we use the other formula.
- $A=C^{-<!--\; -\; -->1}{A}^{\prime }C$ where $A$ is a matrix in a standard basis; $A^{\prime }$ is a matrix in the eigenbasis; $C$ is the transformation matrix.
Why do we have different formulas while doing the same thing (matrix transformations) or do I get something wrong?

For $x\in <!--\; \in \; -->\mathbb{C}$ define $A,B\in <!--\; \in \; -->M(3\times <!--\; \times \; -->3,\mathbb{C})$ as
$A=\left(\begin{array}{ccc}x& 0& 0\\ 0& x& 1\\ 0& 0& x\end{array}\right)$ and $B=\left(\begin{array}{ccc}x& 1& 0\\ 0& x& 0\\ 0& 0& x\end{array}\right).$
How to find a transformation matrix $T\in <!--\; \in \; -->{\text{GL}}_3(\mathbb{C})$ with $B=T^{-<!--\; -\; -->1}AT$

We have $F(x,y)\in <!--\; \in \; -->\mathbb{Z}[X,Y]$ a positive definite binary form of degree $\ge <!--\; \ge \; -->3$. I have to prove, without using lower bounds on linear forms in logarithms (we were working with Baker's theorems), that for each positive integer m the equation $F(x,y)=m$ has only finitely many solutions in $x,y\in <!--\; \in \; -->\mathbb{Z}$. I also have to describe a method to find these.
We know that the coefficient of $X^d$ of $F$ is positive, and that all zeros of $F(X,1)$ are in $\mathbb{C}\mathrm{\setminus <!--\; \setminus \; -->}\mathbb{R}$. We also know the discriminant of $F$ is negative.
I have trouble finding this out by myself. I also looked up some number theory books, but most literature is about binary quadratic forms, not the binary form I have to prove this for. Hope someone can help me!

In linear algebra, to convert the linear transformation or operator into matrix form, it's not hard, get the standard basis of domain and substitute in the transformation and write the image of the standard basis as a linear combination of standard basis of codomain and finally get the constants (c i's) and write that in column wise in the matrix format, we will get the matrix format of the linear transformation with respect to the standard basis of the domain and codomain. My question here is "what about the reverse part?, can we find the linear transformation if the matrix is given?" I have tried multiplying the matrix with the standard basis of Mn(R), but I am getting the columns as the image.. What should I do? To get the transformation.

How can I solve the following exercise?
Let $V$ be the vector space defined on the field $\mathbb{R}$ of $n-<!--\; -\; -->$dimensional. Let $T:V\to <!--\; \to \; -->V$ be a linear transformation such that $T^{n-<!--\; -\; -->1}\ne <!--\; \ne \; -->0$, $T^n=0$ and $B=\{x,T\left(x\right),T^2\left(x\right),\cdots <!--\; \cdots \; -->,T^{n-<!--\; -\; -->1}\left(x\right)\}$ it is a basis for $V$. Find a matrix associated with the linear transformation $T$.
I think the only way to explicitly determine the matrix associated with the linear transformation $T$, is to define $T$ as
$T({\mathrm{\alpha <!--\; \alpha \; -->}}_{1}x+{\mathrm{\alpha <!--\; \alpha \; -->}}_{2}T\left(x\right)+\cdots <!--\; \cdots \; -->+{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}{T}^{n-<!--\; -\; -->1}\left(x\right))={\mathrm{\alpha <!--\; \alpha \; -->}}_{1}x+{\mathrm{\alpha <!--\; \alpha \; -->}}_{2}T\left(x\right)+\cdots <!--\; \cdots \; -->+{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}{T}^{n-<!--\; -\; -->1}\left(x\right)$.

I have a rigid 3D homogeneous transformation matrix 4*4 which includes Rotation (Roll, Yaw and pitch) plus the translation(X,Y and Z).
I want to split this transformation into N small and equal transformations where:
$T_{\text{Original}}=T.T.T.....T$
Is it theoretically OK to just extract the angles and translation values from $T_{\text{Original}}$ and divide them by $N$ then construct the $T$ matrix from them? If not, how can I achieve that?

$T(x,y)=(2x-<!--\; -\; -->y,3x-<!--\; -\; -->y)$
with respect to basis (1,0), and (0,1) Now M(T)=
$\left[\begin{array}{ccc}2& 3& x\\ -<!--\; -\; -->1& -<!--\; -\; -->1& y\end{array}\right]$
gives
$\left(\begin{array}{c}2x+3y\\ -<!--\; -\; -->x-<!--\; -\; -->y\end{array}\right)$
so this cannot be the transformation matrix
The correct one is
$\left[\begin{array}{ccc}2& -<!--\; -\; -->1& x\\ 3& -<!--\; -\; -->1& y\end{array}\right]$
Also, how do I find the matrix transformation of (x,y,z)=(x+y,y+z) for basis (1,0) and (0,1). So we are going from 3 dimensions to 2. So my matrix should have 2 rows and 2 columns right?

How to do this question in regards of matrices and transformation?

From my understanding, this question asks me is there any formula which can make $T(x)=x^2$ a non-linear transformation into a matrix transformation ? What am I supposed to answer to this question?

I would like to solve the following coupled system of linear PDEs by separation of variables, where a and b are constants:
$\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}t}=\frac{b-<!--\; -\; -->a}{a+b}u+(b+a{)}^{2}v+\frac{\mathrm{\partial <!--\; \partial \; -->}{u}^2}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}$
$\frac{\mathrm{\partial <!--\; \partial \; -->}v}{\mathrm{\partial <!--\; \partial \; -->}t}=\frac{-<!--\; -\; -->2b}{a+b}u+-<!--\; -\; -->(b+a{)}^{2}v+d\frac{\mathrm{\partial <!--\; \partial \; -->}{v}^2}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}$
Or, in matrix form:
${\stackrel{\to <!--\; \to \; -->}{w}}_t=A\stackrel{\to <!--\; \to \; -->}{w}+D{\stackrel{\to <!--\; \to \; -->}{w}}_{xx}$
where $\stackrel{\to <!--\; \to \; -->}{w}=\left(\begin{array}{c}u\\ v\end{array}\right),A=\left(\begin{array}{cc}\frac{b-<!--\; -\; -->a}{a+b}& (b+a{)}^2\\ \frac{-<!--\; -\; -->2b}{a+b}& -<!--\; -\; -->(b+a{)}^2\end{array}\right),D=\left(\begin{array}{cc}1& 0\\ 0& d\end{array}\right)$
Since this is a linear equation, one should be able to apply the method of Separation of Variables. I've gone through the exercise of separating variables for single variable models in one or more spatial dimensions many times. However, I've never seen a completely worked example where there are two coupled equations. My intuition tells me to start with:
$\stackrel{\to <!--\; \to \; -->}{w}=\left(\begin{array}{c}u(x,t)\\ v(x,t)\end{array}\right)=\left(\begin{array}{c}{\mathrm{\varphi <!--\; \varphi \; -->}}_1(x)h_1(t)\\ {\mathrm{\varphi <!--\; \varphi \; -->}}_2(x)h_2(t)\end{array}\right)$ Differentiating with respect to x and t, I obtain:
$\left(\begin{array}{c}{\mathrm{\varphi <!--\; \varphi \; -->}}_1(x)h_1^{\prime }(t)\\ {\mathrm{\varphi <!--\; \varphi \; -->}}_2(x)h_2^{\prime }(t)\end{array}\right)=A\left(\begin{array}{c}{\mathrm{\varphi <!--\; \varphi \; -->}}_1(x)h_1(t)\\ {\mathrm{\varphi <!--\; \varphi \; -->}}_2(x)h_2(t)\end{array}\right)+D\left(\begin{array}{c}{\mathrm{\varphi <!--\; \varphi \; -->}}_1^{″}(x)h_1(t)\\ {\mathrm{\varphi <!--\; \varphi \; -->}}_2^{″}(x)h_2(t)\end{array}\right)$
But at this point I get stuck, because expanding this seems to make the problem more difficult.
Textbooks/papers in mathematical biology solve this problem, which is the linearized form of a reaction-diffusion model, by immediately assuming particular solutions of the form:
$\stackrel{\to <!--\; \to \; -->}{w}=\left(\begin{array}{c}u(x,t)\\ v(x,t)\end{array}\right)=\left(\begin{array}{c}{\mathrm{\alpha <!--\; \alpha \; -->}}_1\\ {\mathrm{\alpha <!--\; \alpha \; -->}}_2\end{array}\right)cos(kx)e^{\mathrm{\lambda <!--\; \lambda \; -->}t}$, where k is the wavenumber and $\mathrm{\lambda <!--\; \lambda \; -->}$ is the eigenvalue.
I find this really unsatisfying, because it seems to fall out of nowhere.
My questions are:
1. Is there a general method for separating variables in coupled linear PDEs or PDEs written in vector form?
2. Is there a book/paper/tutorial that I can use to help me work this out in all of its details?
3. Is there some deeper theory that I need to first understand?

Assuming we have a transformation $T(x)=A\ast <!--\; \ast \; -->x$ what is the process for finding the inverse of the transformation $T^{-<!--\; -\; -->1}(x)$?
For example take
$A=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)$

Let $T$ be the transformation of 2 by 2 real symmetric matrices defined by:
$\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]$
$\left[\begin{array}{cc}c& -<!--\; -\; -->b\\ -<!--\; -\; -->b& a\end{array}\right]$
then which of the following statements is NOT true?
1. $det(T)=-<!--\; -\; -->1$
2. $T^{-<!--\; -\; -->1}=T$
3. $T$ is linear
4. the space of 2 by 2 real symmetric matricies with only zeros in the main diagonal is an eigenspace of $T$.
5. $\mathrm{\lambda <!--\; \lambda \; -->}=2$ is an eigen value of $T$

Acos 90 degree matrix transformation.
I'm writing a program that transforms a matrix of points by 90°. In it, I have two vectors from which I am performing the rotation. Both vectors are normalized:
$$\begin{gathered} A:x:\sqrt{(}\frac{1}{3}),y:\sqrt{(}\frac{1}{3}),z:-<!--\; -\; -->\sqrt{(}\frac{1}{3}) \\ B:x:\sqrt{(}\frac{1}{3}),y:\sqrt{(}\frac{1}{3}),z:\sqrt{(}\frac{1}{3}) \end{gathered}$$
As I visualize it, these two vectors are separated by 90°, but the dot product of these vectors comes out to $\frac{1}{3}$
$$\begin{gathered} \sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->\sqrt{(}\frac{1}{3})+\sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->\sqrt{(}\frac{1}{3})+\sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->-<!--\; -\; -->\sqrt{(}\frac{1}{3}) \\ =\frac{1}{3}+\frac{1}{3}-<!--\; -\; -->\frac{1}{3} \\ =\frac{1}{3} \end{gathered}$$
My code is then supposed to use arc-cos to come up with 90° from this number, but I believe arc-cos needs an input of 0 in order to produce a result of 90°. What am I missing here?

Let's say I have a generalised linear (system of) equations. What I mean by that is, say I have something like:
$\underset{i}{\sum <!--\; \sum \; -->}{A}_{i}X{B}_i=C$
Where $A_i$ and $B_i$ are matrices and the equation is to be solved for $X$. Is there a closed-form solution (or failing that, least-squares solution)? As an example, let's take the equation $A_{1}X{B}_1+2aIXI+IX{B}_3=C$. The identity matrices are added to be compatible with the notation I defined above.