Simplifying Linear Algebra: Techniques, Solutions, and Examples

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.
$M=\left(\begin{array}{cccc}0& 0& 1& 5\\ 0& 3& 0& 3\\ -<!--\; -\; -->1& 0& 0& 2\\ 0& 0& 0& 1\end{array}\right)$

Given an arbitrary system of equations, why is direction in space "stored" in the variables when considering the system as linear equations, but "stored" in vectors when considering the system as a vector equation? For example suppose we have a system of three equations in three variables where each equation is of the form $(a_i)x+(b_i)y+(c_i)z=d_i$. Lets also suppose they represent three distinct planes in $R^3$. In the context of the system representing planes in space, it seems to me that dimension/direction is sort of "stored" in the variables $x,y$, and $z$. Considering the system as a linear combination of vectors, the coefficients associated with any one variable make a column vector; for example, for variable $x$ lets call the vector $v_1=$. The vector equation associated with the system would then be $x{v}_1+y{v}_2+z{v}_3=$. In this context it seems as though dimension/direction is stored in the vectors, and the variables $x,y$, and $z$ now just scale them. I realize that both contexts have the same solution set, and both take place in $R^3$, but is there a more intuitive explanation for this relatedness?

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

I wish to find the functional whose minimisation yields the follwoing equation on the vector function u $(\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->})\mathrm{\nabla <!--\; \nabla \; -->}(\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->u)+\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}^{2}u=0$, the Navier equation of linear elasticity. I know that this equation has a vaiational principle from physical reasons, but struggle to find it. The term containing the laplacian is easier to handle, I can not even prove simmetry of the weak form for the first one: how to integrate by parts (grad div u) v to obtain a term symmetric in u e v?

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?

I have to convert the equation $y^2+xy+y=x^3$ by a change of linear variables to the form $Y^2=X^3+aX+b$ where $a$ and $b$ are rational numbers. So far, by completing the square method I've reduced it to
$Y^2=x^3+x^2/4+x/2+1/4$
where $Y=y+(x+1)/2$. However, I can't figure out how to reduce it to the form asked in the question. Any help would be appreciated thanks.

If we have a system of linear equations $Ax=b$, then in case $A$ is invertible it easy to say that the solution is $x=A^{-<!--\; -\; -->1}b$. In all other cases, even if $A$ is invertible, the solution if it exist will be in the form $x=A^{+}b+(I-<!--\; -\; -->A^{+}A)w$, where $A^{+}$ is called pseudoinvers of $A$, and w is a vector of free parameters. My question here is how to compute $A^{+}$ for any matrix $A$? what is the way for doing that? It is easy to compute $A^{+}$ if the column are linearly independent (so that $m\ge <!--\; \ge \; -->n$, $A^{+}=(A^{T}A{)}^{-<!--\; -\; -->1}A$, and also if the rows are linearly independent (so that $m\le <!--\; \le \; -->n$), $A^{+}=A^T(A{A}^T{)}^{-<!--\; -\; -->1}$, but I dont know how to compute $A^{+}$ if $A$ is sequare non invertible matrix or if the columns or rows are not linearly independents. The above does not works. If anyone have any idea about computing pseudoinverse or if there is easier method for finding the solution $x$ for system of linear equations $Ax=b$, please help me in all cases, and please keep in mind that I want to find the general form for the solution if it exist not just for a particular $b$.