Matrix Transformation Examples and Practice Problems

linear transformation from $R^n$ to $R^m$ can be represented in a matrix form. What about a transformation from a:
1) Infinite dimension to infinte dimension
2) finite to infinite dimension
3) infinite to finite dimension Can they represented by matrix form....? Before this another question is there a linear transformation from Infinite dimension to infinte dimension, finite to infinite dimension and vice versa..?

Question: For set of vectors $x_1,x_2$, $x_1=(1,3{)}^T,x_2=(4,6{)}^T$ are in $R^2$. Find the matrix of linear transformation $T:R^2\to <!--\; \to \; -->R^3$ such that $T{x}_1=(-<!--\; -\; -->2,2,-<!--\; -\; -->7{)}^T$ and $T{x}_2=(-<!--\; -\; -->2,-<!--\; -\; -->4,-<!--\; -\; -->10{)}^T$
My attempt: So I'm guessing there exists a matrix (a transformation matrix) and it must be of order (2,3) for it to give (-2,2,-7)${}^T$ when multiplied by (1,3)${}^T$:
$\left[\begin{array}{cc}{t}_{11}& t_{12}\\ t_{21}& t_{22}\\ t_{31}& t_{32}\end{array}\right]$ $\left[\begin{array}{c}1\\ 3\end{array}\right]$ = $\left[\begin{array}{c}-<!--\; -\; -->2\\ 2\\ -<!--\; -\; -->7\end{array}\right]$
But how am I supposed to find out all those ${}_{ij}$ variables? Also, using same on second set of vectors give me inconsistent equations:
$\begin{array}{cc}{t}_{11}+3t_{12}=-<!--\; -\; -->2& 4t_{11}+6t_{12}=-<!--\; -\; -->2\\ t_{21}+3t_{22}=2& 4t_{21}+6t_{22}=-<!--\; -\; -->4\\ t_{31}+3t_{32}=-<!--\; -\; -->7& 4t_{31}+6t_{32}=-<!--\; -\; -->10\end{array}$

Let $ß=(b_1,...,b_n)$ be a basis of the vector space $V$, let $T:V\to <!--\; \backslash rightarrow\; -->V$ be a linear transformation of $V$, and let B be the $ß$-matrix of $T$.
(a) Prove that $v\in <!--\; \in \; -->\ker <!--\; \; -->(T)$ if and only if $[v{]}_B\in <!--\; \in \; -->\ker <!--\; \; -->(B)$.
(b) Prove that $v\in <!--\; \in \; -->\text{Im(T)}$ if and only if $[v{]}_B\in <!--\; \in \; -->\text{Im(T)}$.
(c) Prove that $T$ is an isomorphism if and only if $B$ is invertible.

Given a matrix transformation $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, why does it go from ${\mathbb{R}}^2$ to ${\mathbb{R}}^2$, when it squishes the plane into a line?

Find a matrix for the linear transformation $T:{\mathbb{R}}^3\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^3$ that first rotates a vector by an angle of 30 deegres in the counterclockwise direction about the y-axis, then reflects the vector about the xz-plane, and lastly projects the vector onto the yz-plane.
How would you illustrate a matrix transformation reflecting about the xz-plane in an ${\mathbb{R}}^3$ space?

Find the matrix $A$ of a linear transformation $T:{\mathbb{R}}^2\to <!--\; \backslash rightarrow\; -->{\mathbb{R}}^2$ that satisfies
$T\left[(\genfrac{}{}{0ex}{}{2}{3})\right]=(\genfrac{}{}{0ex}{}{1}{1}),T^2\left[(\genfrac{}{}{0ex}{}{2}{3})\right]=(\genfrac{}{}{0ex}{}{1}{2}).$

Linear transformation matrix derivation
$A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\in <!--\; \in \; -->{\mathbb{R}}^{2\times <!--\; \times \; -->2}$
$L:{\mathbb{R}}^{2\times <!--\; \times \; -->2}⟶<!--\; ⟶\; -->{\mathbb{R}}^{2\times <!--\; \times \; -->2};X\mapsto <!--\; \mapsto \; -->AX$
Find the transformation matrix with respect to the basis
${\mathcal{B}}_1=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],{\mathcal{B}}_2=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],{\mathcal{B}}_3=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],{\mathcal{B}}_4=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$

I have a question regarding matrix transformations. I am trying to transform a parallelogram into the unit square. I need to find a series of matrices which transforms each point.
For example, how do I find a, b, c and d (matrix) when (0,1) is part of the unit square and (3,10) is the original point?
[a b] [3 ] = [0] [c d] [10]

Househbolder transformation identity matrix dimensions
When performing a householder transformation and generating an elementary reflector matrix of the form:
$H=I-<!--\; -\; -->2{\displaystyle \frac{v{v}^T}{v^{T}v}}$
How do we know the dimensions of the identity matrix?

$T\left(\right[\begin{array}{cc}a& b\\ c& d\end{array}\left]\right)=3a-<!--\; -\; -->4b+c-<!--\; -\; -->d$
For a transformation such as this, how does one determine the form of the kernel? Is it simply making the right side equal to zero, solving for each individual variable, and then creating a matrix with these 'new' a,b,c,d values?

my problem came about when applying a 2x2 matrix to a hyperbola..I'm fine with transforming point what I have trouble with is transforming lines.
Theres two types I'd like help with:
(A) : Given a line set to a constant...like y = a, how to I apply this to a matrix? I was thinking of making a column vector with the top part set to zero and the bottom set to a and applying this to the transformation matrix.
(B) : For that hyperbola example, it was a simple unit hyperbola with its asymptotes : y=x, y = -x.
How do I intepret what these transformations do to lines, with respect to arbitrary constants like y=a and for variables.

Question:Find the $3\times <!--\; \times \; -->3$ matrix A, associated with the linear transformation that projects vectors in ${\mathbb{R}}^{\mathbb{3}}$ (orthogonally) onto the plane $x+y+z=0$.
The matrix:
$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

Let us assume A ,B and C are known affine transformation matrices in homogeneous 2D space.
If it should happen that $C=A^m{B}^n$ or some unknown m,n, is there a way to detect this short of trying all the combinations? If one of A,B is an identity matrix this can be determined by examining the eigenvectors, but is there an equivalent way of doing this for a product such as this?

Consider a vector y containing the y values of some putative linear relation in the form of $y=ax+b$ and matrices A and B:
$A=\left(\begin{array}{cc}{x}_1& 1\\ x_2& 1\\ ...& ...\\ x_i& 1\end{array}\right)$
$B=\left(\begin{array}{c}a\\ b\end{array}\right)$
In very basic linear regression theory it is stated that when the residuals are $r=AB-<!--\; -\; -->y$, then $SSQ=r^{T}r=(AB-<!--\; -\; -->y{)}^T(AB-<!--\; -\; -->y)$, which is logic. However, then we can simplify this to $y^{T}y-<!--\; -\; -->2B^T{A}^{T}y+B^T{A}^{T}AB$. I don't really understand how we get this two times B transposed A transposed y, I would simply get: $y^{T}y-<!--\; -\; -->B^T{A}^{T}y+AB{y}^T+B^T{A}^{T}AB$ instead. When I tried to fill in some numbers for the matrices A B en vector y, I indeed find that $AB{y}^T$ equals $A^T{B}^{T}y$.
What is the reasoning/theory behind this?

If $T\in <!--\; \in \; -->\text{End}(V)$ such that $T(x_1)=2x_1+x_2$ and $T(x_2)=x_1$, and $y_1=4x_1+2x_2$ and $y_2=x_1-<!--\; -\; -->x_2$ , determine the matrix $T$ with respect to the basis $\{x_1,x_2\}$ and with respect to the new basis $\{y_1,y_2\}$. Furthermore, it is possible to find an invertible matrix $P$ such that $P^{-<!--\; -\; -->1}AP=B$, where $A$ is the matrix transformation with respect to the basis $\{x_1,x_2\}$ and $B$ is the matrix transformation with respect to the basis $\{y_1,y_2\}$.
For the first part, I know I need to find some matrix $D=C^{-<!--\; -\; -->1}AC$ , such that $A$ is the transformation matrix with respect to the standard basis, and $C$ is the change of basis matrix, but I am unsure how to construct $C$ and thus $C^{-<!--\; -\; -->1}$. The transformation matrix for $T$ is:
$A=\left[\begin{array}{cc}2& 1\\ 1& 0\end{array}\right]$

Let $T:{\mathbb{R}}^3$ -> ${\mathbb{R}}^2$ be a linear transformation satisfying:
$T([1,0,1])=[2,3],T([2,1,3])=[-<!--\; -\; -->1,0],T([0,0,1])=[3,7]$
Find $T([x,y,z])$

Find matrix of linear transformation with respect to the basis of eigenvectors
The transformation $R^2-<!--\; -\; -->>R^2$ is represented by the matrix $\left(\begin{array}{cc}1& 2\\ 1& 0\end{array}\right)$.

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?

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.