Matrix Transformation Examples and Practice Problems

Find the standard matrix for the following composition in ${\mathbb{R}}^{\mathbb{2}}$:
A reflection about the $x$-axis followed by a rotation of $\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$

Let $\mathrm{\psi <!--\; \psi \; -->}$ be a linear operator on $V$, a vector space of dimension two. How do I show, given $\mathrm{\psi <!--\; \psi \; -->}$ is not a scalar multiple of the identity, that there is a $v$ such that $v,\mathrm{\psi <!--\; \psi \; -->}(v)$ forms a basis, and further how do I write the transformation as a matrix with respect to that matrix? For the matrix, the first column would be 0,1, but I don't know what the second column would be.

Let $B=\{v_1,...,v_n\}$ and $C=\{w_1,...,w_n\}$ be bases to V. Suppose: $w_i=m_{i1}{v}_1+...+m_{in}{v}_n$ for $m_{ij}\in <!--\; \in \; -->F,1\le <!--\; \le \; -->i,j\le <!--\; \le \; -->n$ $M$ is an invertible matrix whose $i,j$ member is $m_{ij}$
need to express the transformation matrix, N, from B to C by M. ($\mathrm{\forall <!--\; \forall \; -->}v\in <!--\; \in \; -->V[v{]}_C=N[v{]}_B$)
Is this true?
$$\begin{gathered} (w_1\dots <!--\; \dots \; -->w_n{)}^{-<!--\; -\; -->1}(v_1\dots <!--\; \dots \; -->v_n)=N \\ ((v_1\dots <!--\; \dots \; -->v_n)M^t{)}^{-<!--\; -\; -->1}(v_1\dots <!--\; \dots \; -->v_n)=N \\ (M^t{)}^{-<!--\; -\; -->1}(v_1\dots <!--\; \dots \; -->v_n{)}^{-<!--\; -\; -->1}(v_1\dots <!--\; \dots \; -->v_n)=N \\ (M^{-<!--\; -\; -->1}{)}^t=N \end{gathered}$$

vector:
$\left[\begin{array}{cccccc}a& b& c& d& e& f\end{array}\right]$
which I would like to convert into a 2D matrix. Let's arbitrarily say I'd like to ravel along the rows first (fill one row before moving to the next).
$\left[\begin{array}{cc}a& b\\ c& d\\ e& f\end{array}\right]$
Is there a series of multiplicative matrix transformations that performs this reshaping, and if so, what is the general name for this operation?

Let $\phi :K[x{]}_{\le <!--\; \le \; -->n}\to <!--\; \backslash rightarrow\; -->K[x{]}_{\le <!--\; \le \; -->n-<!--\; -\; -->1}$ with $\mathrm{\phi <!--\; \phi \; -->}$ the linear transformation defind by $\mathrm{\phi <!--\; \phi \; -->}(f)=f^{\prime }$. Select a base and find the matrix of the linear transformation.
I took the standard basis for grade-n polynomials:
$B=<1,x,x^2,\dots <!--\; \dots \; -->,x^n>,\phi (B)=\phi (1,x,x^2,...,x^n)=(0,2x,...,n{x}^{n-<!--\; -\; -->1})$
So, the matrix is $\left[\begin{array}{c}0\\ 2x\\ \dots <!--\; \dots \; -->\\ n{x}^{n-<!--\; -\; -->1}\end{array}\right]$?

Inverse of transformation matrix
For the following 3D transfromation matrix M, find its inverse. Note that M is a composite matrix built from fundamental geometric affine transformations only. Show the initial transformation sequence of M, invert it, and write down the final inverted matrix of M.
$M=\left(\begin{array}{cccc}0& 0& 1& 5\\ 0& 3& 0& 3\\ -<!--\; -\; -->1& 0& 0& 2\\ 0& 0& 0& 1\end{array}\right)$

Showing that the matrix transformation $T(f)=x\ast <!--\; \ast \; -->f^{\prime }(x)+f^{″}(x)$ is linear

Consider a linear transformation $T:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ satisfying
$T\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)\text{and}T\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)=\left(\begin{array}{c}5\\ 6\\ 7\end{array}\right).$The question told me to find $T\left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)$ which I did but I am wondering how would I find the transformation matrix $T$?

Determine the Linear Transformation Matrix
The Matrix R is below:
$\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$
And i have a Linear Transformation below here below that defines by this below.
The linear Transformation $S:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ is given by $S(x)=R(x)+3x.$
how to get the Matrix $S$ from the definition above.

How to solve matrix transformation of a vector?
Given two non-zero vectors $a=(a_1,a_2)$ and $b=(b_1,b_2)$ we define $\sim <!--\; \backslash sim\; -->(a,b)$ by $\sim <!--\; \backslash sim\; -->(a,b)=a_1{b}_2-<!--\; -\; -->a_2{b}_1$. Let $A$, $B$ and $C$ be points with position vectors a, b and c respectively, none of which are parallel. Let $P$, $Q$ and $R$ be points with position vectors $p$, $q$ and $r$ respectively, none of which are parallel.
(i) Show that there exist $2\times <!--\; \times \; -->2$ matrix $M$ such that $P$ and $Q$ are the images of $A$ and $B$ under transformation represented by $M$.

I know that the matrix representation of the linear transformation:
$x_1=X_1+2\mathrm{\lambda <!--\; \lambda \; -->}{X}_2$
$x_2=X_2-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}{X}_1$
is:
$\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{cc}1& 2\mathrm{\lambda <!--\; \lambda \; -->}\\ -<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}& 1\end{array}\right]\left[\begin{array}{c}{X}_1\\ X_2\end{array}\right]$
but what if I have the non-linear transformation:
$x_1=X_1+2\mathrm{\lambda <!--\; \lambda \; -->}{X}_2{X}_1^2$
$x_2=X_2-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}{X}_1$
How can it be expressed in matrix form to get it's inverse?

For each $m\in <!--\; \in \; -->\{1,2,...,n\}$ , is there a transformation ${\mathrm{\varphi <!--\; \varphi \; -->}}_m$ that I can apply to a matrix $M\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ such that the $m^{th}$ largest eigenvalue ${\mathrm{\lambda <!--\; \lambda \; -->}}_m$ of $M$ is the smallest eigenvalue ${\mathrm{\lambda <!--\; \lambda \; -->}}_n^{\prime }$ of the matrix $M^{\prime }={\mathrm{\varphi <!--\; \varphi \; -->}}_m(M)$?

Here's the question:
Let $A=\left[\begin{array}{cc}5& -<!--\; -\; -->3\\ 2& -<!--\; -\; -->2\end{array}\right]$, which is a linear transformation ${\mathbb{R}}^2\to <!--\; \to \; -->{\mathbb{R}}^2$. Find the matrix representing the transformation with respect to basis $\left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]$.
I understand how to solve the question. You would do $B^{-<!--\; -\; -->1}AB$. I did the multiplication and got $\left[\begin{array}{cc}20& 0\\ 0& -<!--\; -\; -->5\end{array}\right]$. The author of the document got the same answer but then multiplied the matrix by 1/5 to get $\left[\begin{array}{cc}4& 0\\ 0& -<!--\; -\; -->1\end{array}\right]$. Why is he allowed to do that?

Suppose that $T:{\mathbb{R}}^n\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^m$ is a linear transformation. Prove that there exists an $m\times <!--\; \times \; -->n$ matrix $A$ such that for all $x\in <!--\; \in \; -->{\mathbb{R}}^n$, $T(x)=Ax$. In other words, $T$ is the "matrix transformation" associated with $A$.

In Sakurai's Modern quantum mechanics it is said that the rotation matrix in three dimensions that changes one set of unit basis vectors $(x,y,z)$ into another set $(x^{\prime },y^{\prime },z^{\prime })$ can be written as
$\left[\begin{array}{ccc}x{x}^{\prime }& x{y}^{\prime }& x{z}^{\prime }\\ y{x}^{\prime }& y{y}^{\prime }& y{z}^{\prime }\\ z{x}^{\prime }& z{y}^{\prime }& z{z}^{\prime }\end{array}\right]$
But shouldn't it be the transpose of matrix given above as the transformation matrix is given by coordinates of transformations of bases ?

Let $X$ be a $m\times <!--\; \times \; -->n$ (m: number of records, and n: number of attributes) normalized dataset (between 0 and 1). Denote $Y=XR$, where $R$ is an $n\times <!--\; \times \; -->p$ matrix, and $p<n$. I understand if $R$ was drawn randomly from Gaussian distribution, e.g.,$N(0,1)$ then the transformation preserve the Euclidean distances between instances (all of the pairwise distances between the points in the feature space will be preserved). But what if $R\sim <!--\; \backslash sim\; -->U(0,1)$, does the transformation still preserve the distance between instances?

Which of the following statements are true?
(1) For $\mathrm{\theta <!--\; \theta \; -->}\mathrm{ϵ<!--\; ϵ\; -->}\mathbb{R}$ fixed, $R:{\mathbb{R}}^{\mathbb{2}}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^{\mathbb{2}},R(x,y)=(x\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->y\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}),x\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+y\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))$ is linear.
(2) The transformation $T:{\mathbb{R}}^{\mathbb{2}}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^{\mathbb{2}},T(x,y)=(1+x^2,1+y^2)$ is linear.
(3) For $n\ge <!--\; \ge \; -->1$, every linear transformation $L:R:{\mathbb{R}}^n\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^n$ satisties $L(\mathbf{0})=\mathbf{0}$
(4) If $L:{\mathbb{R}}^{\mathbb{3}}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^{\mathbb{2}}$ $L:{\mathbb{R}}^{\mathbb{3}}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^{\mathbb{2}}$ is linear, then the matrix which represents $L$ with respect to the standard coordinates must have positive nullity.
(5) If $L:{\mathbb{R}}^{\mathbb{2}}\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^{\mathbb{3}}$ is linear, then the matrix which represents $L$ with respect to the standard coordinates must have positive nullity.
(A) Only (1),(3), and (5) are true.
(B) Only (1),(3), and (4) are true.
(C) Only (2),(3), and (4) are true.
(D) Only (2),(3), and (5) are true.
(E) Only (1),(2), and (5) are true.
I know that (1) is true, (2) is false and (3) is true. What is the difference between (4) and (5) though? How do I know which one is correct? Apparently statement (4) is the correct one but why is that?

Let $a=(a1,a2,a3)$ be a fixed vector in ${\mathbb{R}}^3$. Define the cross product $a\times <!--\; \times \; -->v$ of $a$ and another vector $v=(v_1,v_2,v_3)\in <!--\; \in \; -->{\mathbb{R}}^3$ as
$a\times <!--\; \times \; -->v=det\left[\begin{array}{ccc}{e}_1& e_2& e_3\\ a_1& a_2& a_3\\ v_1& v_2& v_3\end{array}\right]$
Define a function $T:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ by $T(v)=a\times <!--\; \times \; -->v$ for $v\in <!--\; \in \; -->{\mathbb{R}}^3$.
a) Show that $T$ is a matrix transformation and calculate its representing matrix $M$
b) Find $\ker <!--\; \; -->(T)$ and interpret its answer geometrically
c) find $\mathrm{range}<!--\; \; -->(T)$ and interpret its answer geometrically

A $n\times <!--\; \times \; -->n$ matrix $A$ is a linear transformation from the space of $n\times <!--\; \times \; -->n$ matrices to itself. What is the $n^2\times <!--\; \times \; -->n^2$ matrix corresponding to this transformation? If we knew the nullity and rank of $A$, how could we find the nullity and rank for this transformation?

The transformation matrix $P$ represents a ${90}^{\circ <!--\; \circ \; -->}$ anticlockwise rotation about the origin.
Describe fully the single transformation represented by the matrix $P^3$
From my working, P would equate to
$P=\left[\begin{array}{cc}0& -<!--\; -\; -->1\\ 1& 0\end{array}\right]$
Now to find the transformation shown by $P^3$ would I have to cube the matrix or is there something that I'm missing here?