Simplifying Linear Algebra: Techniques, Solutions, and Examples

Define a matrix transformation A as follows:
A = {{1, -1}, {2, -2}, {3, -3}};
Find the image of {1,3} under A and determine if {3,6,9} is in the range of the transformation.

Reflect a point about a plane using matrix transformation
Given the plane equation $ax+by+cz=d$ and an arbitrary point in 3D space $(i,j,k)$, how do I find a homogenous transformation matrix that is used to reflect this point about the plane?

I'm attempting one of the optional problems in my linear algebra textbook and I am very confused as I have never seen a problem in this form. Question is:
Find the vector form of the equation of the line in ${\mathbb{R}}^3$ that passes through the point $P=(-<!--\; -\; -->1,1,3)$ and is perpendicular to the plane with general equation $x-<!--\; -\; -->3y+2z=5$.
I know the normal vector $\stackrel{\to <!--\; \to \; -->}{n}=[1,-<!--\; -\; -->3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?
My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\stackrel{\to <!--\; \to \; -->}{u}=\stackrel{\to <!--\; \to \; -->}{P}Q$ and $\stackrel{\to <!--\; \to \; -->}{v}=\stackrel{\to <!--\; \to \; -->}{P}R$. Then I would use those to form my vector equation
$\stackrel{\to <!--\; \to \; -->}{x}=\stackrel{\to <!--\; \to \; -->}{p}+s\stackrel{\to <!--\; \to \; -->}{u}+t\stackrel{\to <!--\; \to \; -->}{v}$
Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.
I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.

Linear transformation: $T:\mathbb{R}[X{]}_{\le 2}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$
With $T(f):=(f(0),f^{\prime }(1),f(2))$ and $\mathbb{R}[X{]}_{\le 2}$ is the $\mathbb{R}$-vector space of the polynomials of degree ≤2
Portray the linear transformation T as matrix referring to the basis $1,X,X^2$ of $\mathbb{R}[X{]}_{\le 2}$ and the standard basis of ${\mathbb{R}}^3$
Now I've done the transformation for both the bases, but I don't know if that works:
For the first one I obtain a matrix:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 2\\ 1& 2& 4\end{array}\right]$
Doing $T(1)=(1,0,1)$, $T(X)=(0,1,2)$, $T(X^2)=(0,2,4)$
And for the standard basis I obtained a matrix:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$
Doing the same procedure with $T(e_1),T(e_2),T(e_3)$
I'm not sure it's right, and I don't know if I have to get just one another matrix instead of one.

If A is not symmetric, then A represwnts a non-uniform scalling, followed by an anti-clockwise rotation by an angle of $\mathrm{\theta <!--\; \theta \; -->}=\arctan <!--\; \; -->(\frac{c-<!--\; -\; -->b}{a+d})$
$\left[\begin{array}{cc}0& -<!--\; -\; -->2\\ 4& 0\end{array}\right]$

Let $M_2(\mathbb{R})$ the vector space generated by all the square matrices of $2\times <!--\; \times \; -->2$. Consider the linear transformation $T:<!--\; :\; -->M_2(\mathbb{R})⟶<!--\; ⟶\; -->M_2(\mathbb{R})$ given by $T(A)=A^T$ (where $A^T$ is the transpose of $A$). Calculate a basis for $M_2(\mathbb{R})$ such that the transformation $T$ is represented by a diagonal matrix. Which are the possible values for the diagonal?

How do I use matrix multiplication to find the reflection of (-1,2) about the x axis, y axis and the line y=x?

Does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$. The answer to this is obviously no as I can vary the dimension of $M$. But now this lead me to think , if I take , lets say only $2\times <!--\; \times \; -->2$ matrix into consideration. Now for a matrix $M$, $A=M^T{M}^{-<!--\; -\; -->1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and I had read that any linear transformation can be represented as a matrix. So is the last statement wrong or my argument flawed?

Finding alternate transformation matrix for similarity transformation
A pair of square matrices $X$ and $Y$ are called similar if there exists a nonsingular matrix $T$ such that $T^{-<!--\; -\; -->1}XT=Y$ holds. It is known that the transformation matrix $T$ is not unique for given $X$ and $Y$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...
What I want to mean is: Given $X$ and $Y$ a pair of similar matrices, if $S$ and $T$ are two possible transformation matrices satisfying $S^{-<!--\; -\; -->1}XS=T^{-<!--\; -\; -->1}XT=Y$, is there any generic (apart from scaling) relation between $T$ and $S$ (e.g., direction of column vectors)?
For a specific example, consider $X=\left[\begin{array}{cc}A& BK\\ C& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}A+A^{-<!--\; -\; -->1}BKC& -<!--\; -\; -->A^{-<!--\; -\; -->1}BKC{A}^{-<!--\; -\; -->1}B\\ KC& -<!--\; -\; -->KC{A}^{-<!--\; -\; -->1}B\end{array}\right]$. Assuming $K$ to be invertible it can be shown that $X$ and $Y$ are similar with transformation matrix $T=\left[\begin{array}{cc}I& -<!--\; -\; -->A^{-<!--\; -\; -->1}B\\ 0& K^{-<!--\; -\; -->1}\end{array}\right]$. Can there be any other matrix $S$ which will be independent of $K$, and would result $S^{-<!--\; -\; -->1}XS=Y$?

What is the general transformation matrix for a rotation of angle $\mathrm{\theta <!--\; \theta \; -->}$ about the origin?
That is all the questions says, any one able to help me out, who may understand it better then me?

The form $f(x)y^n(x)+...u(x)y(x)=h(x)$ is supposed to be the definition of lineaity in diff equations. It excludes functions of y in the right hand side, but is multiplication of y by another function allowed in the right hand side allowed? It seems to be the case sometimes but not all of the time, as only composition of linear functions gives linear functions. Can't we just move it to the other side with the other y and call that form linear as well?

Let $f:{\mathbb{R}}^m\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^n$ be a linear transformation. As is common knowledge, it can be expressed as an $n\times <!--\; \times \; -->m$ matrix.

I have a non-linear form of the Poisson equation (with a diffusion coefficient that is a function of the derivatives of the dependent variable) that I'm trying to solve numerically (using FEM which requires the pde to be posed in its weak form).
$\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->(\mathrm{\eta <!--\; \eta \; -->}\mathrm{\nabla <!--\; \nabla \; -->}v)=G(x)$
where $\mathrm{\eta <!--\; \eta \; -->}=\sqrt{1/2(\mathrm{\nabla <!--\; \nabla \; -->}v:\mathrm{\nabla <!--\; \nabla \; -->}v)}$ and $v=v(y,z)$. Multiplying by a test function θ and integrating wrt the domain, $\mathrm{\Omega <!--\; \Omega \; -->}$, gives the weak form
${\oint <!--\; \oint \; -->}_{\mathrm{\Gamma <!--\; \Gamma \; -->}}\mathrm{\theta <!--\; \theta \; -->}\cdot <!--\; \cdot \; -->(\mathrm{\eta <!--\; \eta \; -->}\mathrm{\nabla <!--\; \nabla \; -->}v\cdot <!--\; \cdot \; -->\mathbf{n})d\mathrm{\Gamma <!--\; \Gamma \; -->}-<!--\; -\; -->{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\eta <!--\; \eta \; -->}\mathrm{\nabla <!--\; \nabla \; -->}v\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathrm{\theta <!--\; \theta \; -->}+G)d\mathrm{\Omega <!--\; \Omega \; -->}=0$
However, none of my boundary conditions correspond to the Dirichelet type as I have ${\mathrm{\partial <!--\; \partial \; -->}}_{n}v=v$ on three of the boundaries in a rectangular domain and ${\mathrm{\partial <!--\; \partial \; -->}}_{n}v=0$ on the other (n is the normal vector to surface). Is this problem actually solvable using FEM, or at all? I have the solution to a simpler problem which might be a reasonable guess at the solution here.

I am studying about the linear odes with non-constant coefficients.
I know the first order linear ode with non-constant coefficient
$\begin{array}{}\text{(1)}& y^{{}^{\prime }}(x)+f(x)y(x)=0\end{array}$
has a general solution of the form
$\begin{array}{}\text{(2)}& y=C{e}^{-<!--\; -\; -->\int <!--\; \int \; -->f(x)dx}\end{array}$
However, I am more interested in the case of linear second order odes with non-constant coefficients
$\begin{array}{}\text{(3)}& y^{{}^{″}}(x)+g(x)y^{{}^{\prime }}(x)+f(x)y(x)=0\end{array}$
I know that this equation does not have a closed form solution like (2). However, I am interested in special cases of that.
Questions
1. Consider (3), when $g(x)=0$, then we have
$\begin{array}{}\text{(4)}& y^{{}^{″}}(x)+f(x)y(x)=0\end{array}$
Is Eq.(4) a famous well-known equation? If YES, what is its name?
2. Does (4) have a closed form solution like (2)?
3. Can you name or give me a list of well-known linear second order odes with non-constant coefficients which are not polynomial?
For example, I know Cauchy-Euler, Airy, Bessel, Chebyshev, Laguerre and Legendre equations whose coefficients are polynomials. But I don't know any well-known equation with non-polynomial coefficients.

Find linear transformation matrix from a linear transformation matrix of its composition
Let's say that $V$ is a vector space over the field $K$ and suppose we have a linear transformation $f\in <!--\; \in \; -->\text{End}V$ thats matrix is know on some basis.
How to find a matrix of a linear transformation $g\in <!--\; \in \; -->\text{End}V$ on the same basis, so that $g\circ <!--\; \circ \; -->g=f$.
For example if $V$ is vector space over the field ${\mathbb{Z}}_{11}$ and matrix of linear transformation $f\in <!--\; \in \; -->\text{End}V$ on some basis is
$\left(\begin{array}{cccc}\stackrel{¯<!--\; ¯\; -->}{7}& \stackrel{¯<!--\; ¯\; -->}{3}& \stackrel{¯<!--\; ¯\; -->}{4}& \stackrel{¯<!--\; ¯\; -->}{0}\\ \stackrel{¯<!--\; ¯\; -->}{0}& \stackrel{¯<!--\; ¯\; -->}{6}& \stackrel{¯<!--\; ¯\; -->}{9}& \stackrel{¯<!--\; ¯\; -->}{6}\\ \stackrel{¯<!--\; ¯\; -->}{1}& \stackrel{¯<!--\; ¯\; -->}{9}& \stackrel{¯<!--\; ¯\; -->}{3}& \stackrel{¯<!--\; ¯\; -->}{1}\\ \stackrel{¯<!--\; ¯\; -->}{0}& \stackrel{¯<!--\; ¯\; -->}{2}& \stackrel{¯<!--\; ¯\; -->}{8}& \stackrel{¯<!--\; ¯\; -->}{5}\end{array}\right)$
Find matrix of a linear transformation $g\in <!--\; \in \; -->\text{End}V$ on the same basis, so that $g\circ <!--\; \circ \; -->g=f$

Find covariance of matrix transformation
$VX=\left[\begin{array}{ccc}4& & 4\\ 4& & 16\end{array}\right]$
If $Y_1=X_1+2$ and $Y_2=X_2-<!--\; -\; -->X_1+1$, then how do I find the variance matrix for $Y$?
I have tried the following where I emitted the constants as my guess is they don't affect variance:
$Cov(Y_1,Y1)=Cov(X_1,X_1)=Var(X_1)=4$
$$\begin{gathered} Cov(Y_2,Y_2)=Cov(X_2-<!--\; -\; -->X_1,X_2-<!--\; -\; -->X_1) \\ =Cov(X_2,X_2-<!--\; -\; -->X_1)-<!--\; -\; -->Cov(X_1,X_2-<!--\; -\; -->X_1) \\ =Cov(X_2,X_2)-<!--\; -\; -->Cov(X_2,X_1)-<!--\; -\; -->Cov(X_1,X_2)+Cov(X_1,X_1) \\ =16-<!--\; -\; -->4-<!--\; -\; -->4+4=12 \end{gathered}$$
Variance = $\left[\begin{array}{ccc}4& & 0\\ 0& & 12\end{array}\right]$

I am trying to show that the functions $t^3$ and $b$ are independent on the whole real line. To do this, I try and prove it by contradiction. So assume that they are dependent. So then there must exists constants $a$ and $b$ such that $a{t}^3+b|t{|}^3=0$ for all $t\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->})$. Now pick two points $x$ and $y$ in this interval and assume without loss of generality that $x<0$, $y\ge <!--\; \ge \; -->0$. Now form the simultaneous linear equations
$a{y}^3+b|y{|}^3=0$, viz.
$\left[\begin{array}{cc}{x}^3& |x{|}^3\\ y^3& |y{|}^3\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$
Now here's my problem. If I look at the determinant of the coefficient matrix of this system of linear equations, namely $x^3|y{|}^3-<!--\; -\; -->y^3|x{|}^3$ and noting that $x<0$ and $y>0$, I have that the determinant is non-zero which implies that the only solution is $a=b=0$, i.e. the functions $t^3$ and |$|t{|}^3$ are linearly independent. However what happens if indeed $y=0$? Then the determinant of the matrix is 0 and I have got a problem.
Is there something that I am not getting from the definition of linear independence?
The definition (I hope I state this correctly) is: If $f$ and $g$ are two functions such that the only solution to $af+bg=0\mathrm{\forall <!--\; \forall \; -->}t$ in an interval $I$ is $a=b=0$, then the two functions are linearly independent.
But what happens if my functions pass through the origin, like the above? Then I've just shown that there exists a $t$ in an interval containing zero such that the two functions are zero, viz. I can plug in any $b$ and $a$ such that $af+bg=0$.

I have to find components of a matrix for 3D transformation. I have a first system in which transformations are made by multiplying:
$M_1=[Translation]\times <!--\; \times \; -->[Rotation]\times <!--\; \times \; -->[Scale]$
I want to have the same transformations in an engine who compute like this:
$M_2=[Rotation]\times <!--\; \times \; -->[Translation]\times <!--\; \times \; -->[Scale]$
So when I enter the same values there's a problem due to the inversion of translation and rotation.
How can I compute the values in the last matrix $M_2$ for having the same transformation?

I have a hard time finding the analytical solution to the following non-linear equation:
$(1+x{)}^{1-<!--\; -\; -->p}+p^{\frac{p}{1-<!--\; -\; -->p}}x(p-<!--\; -\; -->1)-<!--\; -\; -->1=0$
where $p\in <!--\; \in \; -->]0,1[$ and $x>1$. I would like to have a solution $x$ in terms of $p$ for each fixed $p$ in the specified interval. Of course it's possible that no analytical form of the solution exists. If so, I'd also be happy to hear an argument why that is the case. However, for some values of p the solution is 'nice', for example for $p=\frac{1}{2}$ it's easy to compute that $x=8$ and for $p=\frac{1}{3}$ I obtained $x=\frac{9}{16}(5\sqrt{3}+\sqrt{11+64\sqrt{3}})$
Any kind of help is greatly appreciated!