Matrix Transformation Examples and Practice Problems

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?

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.

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 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?

Which transformations include in the matrix?
do this matrix transformation options:
1) rotate and scaling
2) translate and scaling
3) scaling and rotate
4) not 1,2,3
$\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]$

Let $T:{\mathbb{R}}^2\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ and $T(-<!--\; -\; -->2,3)=(-<!--\; -\; -->1,0,1)$ and $T(1,2)=(0,-<!--\; -\; -->1,0)$
Obtain the canonical matrix of $T$ and the transformation $T(x,y)$.

Let $\mathrm{\Sigma <!--\; \Sigma \; -->}$ be a symmetric positive definite matrix with ones on the diagonal (= correlation matrix).
Let $A$ be an invertible matrix.
I'm pretty sure that if $\mathrm{\Omega <!--\; \Omega \; -->}:=A\mathrm{\Sigma <!--\; \Sigma \; -->}{A}^t$ has ones on its diagonal, then
$\mathrm{\Omega <!--\; \Omega \; -->}=Id\text{or}\mathrm{\Omega <!--\; \Omega \; -->}=\mathrm{\Sigma <!--\; \Sigma \; -->}$
(which would correspond to $A=Id$ or $A=chol(\mathrm{\Sigma <!--\; \Sigma \; -->})$)
But I don't know how to proove it.

Let $B=\{u_1,u_2,u_3\}$ with $u_1=1$, $u_2=x$, $u_3=x^2$ denote a basis for the space of polynomials of the second order polynomials $P^2$. Let $T(a_0+a_{1}x+a_2{x}^2)=a_0+a_1(x-<!--\; -\; -->1)+a_2(x-<!--\; -\; -->1{)}^2$ be a linear transformation form $P^2$ to itself. Construct the representation matrix $[T{]}_{BB}.$.

We know that if we want to reflect any point over an origin, i.e. $O(0,0)$, we can use matrix transformation like this
$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}-<!--\; -\; -->1& 0\\ 0& -<!--\; -\; -->1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-<!--\; -\; -->x\\ -<!--\; -\; -->y\end{array}\right).$
But, what if we reflect any point over another point $M(a,b)$ with $a,b\ne <!--\; \ne \; -->0$

How to formulate a transformation matrix for the following operation? Like all the examples I found are different and I can't understand how to solve this problem:
$$\begin{gathered} y=A.X \\ y=(y_1{y}_2{)}^{\prime } \\ x=(x_1^{\prime }{x}_2) \\ (y_1{y}_2{)}^{\prime }=(x_1{x}_2{)}^{\prime } \\ (-<!--\; -\; -->x_1,x_2{)}^{\prime }=(-<!--\; -\; -->x_1,-<!--\; -\; -->x_2{)}^{\prime } \end{gathered}$$

following matrix
$\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$
Results in the shape being twice as small.
Now I am trying to use this matrix and apply it somewhere else, but because of the inverse scale I am unable to use it. Is there any way to normalize the scaling factor of a affine transformation matrix?
One last thing to note is the fact that the rotation applied, according to documentation is also reversed from standard formulas. Their rotation matrix would look something like this:
$\left[\begin{array}{cc}cos& sin\\ -<!--\; -\; -->sin& cos\end{array}\right]$

Two 3 dimensional points. $A[x_1,y_1,z_1]$ and $B[x_2,y_2,z_2]$. Need to find a transformation matrix which when multiplied to A will give me B and when multiplied by B give me A. The transformation needs to be a reflection against the plane that's perpendicular to the middle of the AB segment and passing through the midpoint of the AB.

In the vectorspace ${\mathbb{R}}^2$ two vectors are given (Which also form the basis a): $a_1=(8,-<!--\; -\; -->3)$ and $a_2=(5,-<!--\; -\; -->2)$
A linear mapping is determined by: $f(a_1)=2a_1-<!--\; -\; -->4a_2$ and $f(a_2)=-<!--\; -\; -->a_1+2a_2$
How can one determine the transformation matrix of f with respect to the standard e-basis?

For the displacement of the element by n rows from the identity matrix, the element should hold the value $(\frac{1}{2}{)}^n$ in the transformed matrix.
Here are a few examples:
$A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\to <!--\; \backslash rightarrow\; -->A^{\prime }=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
$B=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)\to <!--\; \backslash rightarrow\; -->B^{\prime }=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& \frac{1}{2}\\ 0& \frac{1}{2}& 0\end{array}\right)$
$C=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)\to <!--\; \backslash rightarrow\; -->C^{\prime }=\left(\begin{array}{ccc}0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{2}\\ \frac{1}{4}& 0& 0\end{array}\right)$
How can such a transformation be realized mathematically?