Simplifying Linear Algebra: Techniques, Solutions, and Examples

Let $V$ be the vector space of polynomials of degree up to 2.
and $T:V\to <!--\; \backslash rightarrow\; -->V$ be a linear transformation defined by the type:
$T(p(x))=p(2x+1)$
Find the matrix form of this linear transformation. The base to find the matrix is $B=\{1,x,x^2\}$

Unimodular Matrix is a matrix, which has integer entries and determinant of $\in <!--\; \in \; -->\{+1,-<!--\; -\; -->1\}$. That means the matrix is invertible over the integers $\mathbb{Z}$.
Given a unimodular matrix A for example:
$\left[\begin{array}{cccccc}2& -<!--\; -\; -->2& -<!--\; -\; -->5& -<!--\; -\; -->4& -<!--\; -\; -->2& -<!--\; -\; -->3\\ 0& 1& 0& 0& 0& 0\\ -<!--\; -\; -->1& 0& 3& 1& -<!--\; -\; -->1& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]$
Need to convert to it to a matrix like this:
$\left[\begin{array}{cccccc}\ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->\\ \ast <!--\; \ast \; -->& \ast <!--\; \ast \; -->& 0& 0& 0& 0\\ \ast <!--\; \ast \; -->& 0& \ast <!--\; \ast \; -->& 0& 0& 0\\ \ast <!--\; \ast \; -->& 0& 0& \ast <!--\; \ast \; -->& 0& 0\\ \ast <!--\; \ast \; -->& 0& 0& 0& \ast <!--\; \ast \; -->& 0\\ \ast <!--\; \ast \; -->& 0& 0& 0& 0& \ast <!--\; \ast \; -->\end{array}\right]$
where * denotes possible nonzero entries.

Given the linear system of equations:
$\{\begin{array}{l}{x}_1+x_2+x_3=n\\ x_1+x_2+x_3+x_4+x_5=3n\\ 2x_1+2x_2+2x_3+x_4+x_5+3x_6+3x_7+3x_8=10n\end{array}$
how many solutions are in $\mathbb{N}\cup <!--\; \cup \; -->\{0\}$?
The solution must not be using sum notation like $\sum <!--\; \sum \; -->y$.
I know how to find the number of solutions to the regular equations like $x_1+x_2+x_3+\cdots <!--\; \cdots \; -->=n$ but I'm not sure how to do this for a system of equations. I thought of substituting some $x$'s:
$$\begin{gathered} x_1+x_2+x_3=3n-<!--\; -\; -->(x_4+x_5) \\ {x}_4+x_5=10n-<!--\; -\; -->2(x_1+x_2+x_3)-<!--\; -\; -->3(x_6+x_7+x_8) \\ ⟹<!--\; ⟹\; -->x_1+x_2+x_3=3n-<!--\; -\; -->(10n-<!--\; -\; -->2(x_1+x_2+x_3)-<!--\; -\; -->3(x_6+x_7+x_8)) \\ ⟹<!--\; ⟹\; -->x_1+x_2+x_3+3(x_6+x_7+x_8)=7n\ast <!--\; \ast \; --> \end{gathered}$$
As far as I understand finding the number of solutions for the system is equivalent to finding the number of solutions to the equation *.
The only next step from here I can think of is using generating functions:
$(1+x+x^2+\dots <!--\; \dots \; -->{)}^3(1+x^3+x^6+x^9+\dots <!--\; \dots \; -->{)}^3$
and we need to find the coefficient of $x^{7n}$.
From the closed form identities we have:
$\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{3-<!--\; -\; -->1+k}{k})x^k\cdot <!--\; \cdot \; -->\underset{i=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{3-<!--\; -\; -->1+i}{i})x^{3i}$
But I have no idea now how to find the coefficient of 7n from here and certainly not without using some kind of sum notation.

Let $K=\mathbb{R}$ and $V:=\mathbb{R}[t{]}_2$ and let $F:\mathbb{R}[t{]}_2\to <!--\; \backslash rightarrow\; -->\mathbb{R}[t{]}_2,f\mapsto <!--\; \mapsto \; -->f(0)+f(1)\cdot <!--\; \cdot \; -->(t+t^2)$
Want to calculate $M_S^S(F)$, the transformation Matrix of F with Basis S, being the standardbasis for the polynomialring $S=(1,t,t^2)$
So we have to use the coordinate isomorphism:
$M_S^S(F)=\left(\begin{array}{ccc}{I}_S(F(1))& I_S(F(t))& I_S(F(t^2))\end{array}\right)$
Already know the solution, however I have no idea why the following matrix is the transformation Matrix of F.
$M_S^S(F)=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)$
Looking at the transformation matrix we know that $F(1)=1+t+t^2$?

I have a problem understanding getting the KERNEL and IMAGE of a linear transformation. We have the following transformation given:
${\mathbb{R}}_2[x]\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}_2[x]$
$(\mathrm{\varphi <!--\; \varphi \; -->}(p))(x)=(xp(x+1){)}^{\prime }-<!--\; -\; -->2p(x)$
We first have to find its matrix in basis
$\{1,x,x^2\}$
which I know how to get. The transformation matrix result is:
$\left[\begin{array}{ccc}-<!--\; -\; -->1& 1& 1\\ 0& 0& 4\\ 0& 0& 1\end{array}\right]$
How do I get the KERNEL and the IMAGE from it ?

Let there be a linear transformation going from ${\mathbb{R}}^3$ to ${\mathbb{R}}^2$, defined by $T(x,y,z)=(x+y,2z-<!--\; -\; -->x)$. Find the transformation matrix if base 1:
$⟨<!--\; ⟨\; -->(1,0,-<!--\; -\; -->1),(0,1,1),(1,0,0)⟩<!--\; ⟩\; -->$,
base 2: $⟨<!--\; ⟨\; -->(0,1),(1,1)⟩<!--\; ⟩\; -->$
An attempt at a solution included calculating the transformation on each of the bases in ${\mathbb{R}}^3$, (base 1) and then these vectors, in their column form, combined, serve as the transformation matrix, given the fact they indeed span all of $B_1$ in $B_2$
Another point: if the basis for ${\mathbb{R}}^3$ and ${\mathbb{R}}^2$ are the standard basis for these spaces, the attempt at a solution is a correct answer.

What is the transformation matrix?
In ${\mathbb{R}}^{\mathbb{2}}$ a basis is given $a=(a_1,a_2)$ where:
$a_1=(1,-<!--\; -\; -->1)$
$a_2=(0,1)$
For $f:{\mathbb{R}}^{\mathbb{2}}\to <!--\; \to \; -->{\mathbb{R}}^{\mathbb{2}}$ it is known:
$f(a_1)=-<!--\; -\; -->6\cdot <!--\; \cdot \; -->a_1$
$f(a_2)=1\cdot <!--\; \cdot \; -->a_2$
Determine the matrix of transformation with regards to the standard e-basis.

Similarity transformation of an orthogonal matrix
A transformation $T$ represented by an orthogonal matrix $A$ , so $A^{T}A=I$. This transformation leaves norm unchanged.
I do a basis change using a matrix $B$ which isn't orthogonal , then the form of the transformation changes to $B^{-<!--\; -\; -->1}AB$ in the new basis( A similarity transformation).
Therefore $B^{-<!--\; -\; -->1}AB$. $[B^{-<!--\; -\; -->1}AB{]}^T=I$.
This suggests that $B^{T}B=I$ which means it is orthogonal, but that is a contradiction.

can anyone give me a reference about the following kind of generalised spherical Bessel equation:
$[x^2{\mathrm{\partial <!--\; \partial \; -->}}_x^2+2x{\mathrm{\partial <!--\; \partial \; -->}}_x+k^2(x^2+\mathrm{\alpha <!--\; \alpha \; -->}x)-<!--\; -\; -->l(l+1)]f(x)=0$
or for $u(x)=xf(x)$:
$[{\mathrm{\partial <!--\; \partial \; -->}}_x^2+k^2(1+\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{x})-<!--\; -\; -->\frac{l(l+1)}{x^2}]u(x)=0.$
I could not find any "modified" or "generalised" form of Bessel's equation that included a linear term like the $k^2\mathrm{\alpha <!--\; \alpha \; -->}x$ term above.

a) Give the matrix for the Transformation $T:{\mathbb{R}}^2\to <!--\; \to \; -->{\mathbb{R}}^2$ that first reflects points through the $x$ - axis and then reflects through the line $y=x$
b) How else can you describe this transformation?

Let $V$ denote the real vector space $R^2$ and $\mathrm{\psi <!--\; \psi \; -->}:V\to <!--\; \backslash rightarrow\; -->V$ be a real linear transformation such that $\mathrm{\psi <!--\; \psi \; -->}((1,0))=(11,8)$ and $\mathrm{\psi <!--\; \psi \; -->}((0,1))=(4,3)$. Express the image $\mathrm{\psi <!--\; \psi \; -->}((x,y))$ of $(x,y)$ in terms of $x$ and $y$. Assume that $w_1=(4,5)$ and $w_2=(9,11)$ form an ordered basis $B$ for $V$ . Working from the denition determine the matrix $M_B^B(\mathrm{\psi <!--\; \psi \; -->})$ with respect to the basis $B$.
does $M_B^B(\mathrm{\psi <!--\; \psi \; -->})=\left[\begin{array}{cc}-<!--\; -\; -->469& -<!--\; -\; -->1048\\ 388& 867\end{array}\right]$?

Find the general solution of the equation
$\frac{dy}{dx}=1+xy$
My attempt: Arranging it in standard linear equation form $\frac{dy}{dx}+Py=Q$, we get
$\frac{dy}{dx}+(-<!--\; -\; -->x)y=1$
Hence, the integrating factor(I.F.) = $e^{\int <!--\; \int \; -->-<!--\; -\; -->xdx}=e^{-<!--\; -\; -->x^2/2}$. Hence, the solution is
$y(e^{-<!--\; -\; -->x^2/2})=\int <!--\; \int \; -->e^{-<!--\; -\; -->x^2/2}dx+c$
How do I solve this integral now?

If a nonzero matrix $A$ is transformed from ${\mathbb{R}}^3$ to ${\mathbb{R}}^2$, then the null space of $A$ must be a one dimensional (sub)space of ${\mathbb{R}}^3$.
So i know that null space of $A$ is $\{x:<!--\; :\; -->Ax=0\}$ and I also know the definition of not onto. I don't understand the whole concept of one-dimensional space and would this statement be always true?

Let $T:R_3\to <!--\; \backslash rightarrow\; -->R_3$ be the linear transformation corresponding to rotating by $\mathrm{\pi <!--\; \pi \; -->}/4$ clockwise around the z-axis, and then reflecting over the x-axis. Since T is a linear transformation, it corresponds to left multiplication by some matrix A.

An example asks me to define $T:M_{2\times <!--\; \times \; -->2}(F)\to <!--\; \backslash rightarrow\; -->M_{2\times <!--\; \times \; -->2}$ by $T(A)=A^t$. Compute $[T{]}_{\mathrm{\alpha <!--\; \alpha \; -->}}$
$\mathrm{\alpha <!--\; \alpha \; -->}$ is the standard ordered basis of $2\times <!--\; \times \; -->2$ matrices.
To find the transformation I performed the transformation on all four elements of $\mathrm{\alpha <!--\; \alpha \; -->}$, for example $T(\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right))=(\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right))$ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\times <!--\; \times \; -->4$ matrix:
$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right)$
I don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\mathrm{\alpha <!--\; \alpha \; -->}$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in α since one is a $4\times <!--\; \times \; -->4$ and one is a $2\times <!--\; \times \; -->2$
Also, why doesn't this mean that any arbitrary $2\times <!--\; \times \; -->2$ matrix can be raised to the power of t by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).

Let $\mathcal{B}=\{{\mathbf{b}}_1,{\mathbf{b}}_2\}$ be the basis for ${\mathbb{R}}^2$ with ${\mathbf{b}}_1=\left[\begin{array}{c}1\\ 1\end{array}\right]$, ${\mathbf{b}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$. Furthermore, let $T:{\mathbb{R}}^2\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^2$ be a linear transformation. The matrix representation of $T$ with respect to $\mathcal{B}$ is $[T{]}_{\mathcal{B}}=\left[\begin{array}{cc}1& -<!--\; -\; -->1\\ 1& 2\end{array}\right]$. What is the standard matrix of $T$?