Matrix Transformation Examples and Practice Problems

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 $M_B^B(\mathrm{\psi <!--\; \psi \; -->})=\left[\begin{array}{cc}-<!--\; -\; -->469& -<!--\; -\; -->1048\\ 388& 867\end{array}\right]$?

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 $T$?

Rotate a triangle with vertices (2, 1). (4. 1). (3, 3) by 90 degrees (clock wise) and then Translate it by Tx 4 and Ty 3 and finally rotate it by 30 degrees (anti-clock wise). (cos90=0, sin90 1. cos 30 = 0.866, sin30 =0.5)
T1 is Rotation with 90 T2 is Translation T3 is Rotation with 30
What is correct order of transformation matrix to multiply the metrix?
T3 T2 T1 or T1 T2 T3?
Should i follow the order given by the question or reverse order?

Representing a Linear Transformation as a Matrix
$T(\left[\begin{array}{ccc}a& & b\\ c& & d\end{array}\right])=\left[\begin{array}{ccc}d& & a\\ b& & c\end{array}\right]$
I attempted to apply $T$ to the standard basis of the $2\times <!--\; \times \; -->2$ matrices. However, I ended up with a $4\times <!--\; \times \; -->4$ matrix (understandable, as there are $4$ matrices in that basis). This matrix I came up with was
$\left[\begin{array}{ccccccc}0& & 0& & 0& & 1\\ 1& & 0& & 0& & 0\\ 0& & 1& & 0& & 0\\ 0& & 0& & 1& & 0\end{array}\right]$

representative matrix of a linear transformation
Let $T:{\mathbb{F}}^2\to <!--\; \to \; -->{\mathbb{F}}^2$ a linear tranformation such that $T^2=0$, Prove that $T=0$, or there exists a basis of $M_2(\mathbb{F})$ in which the representative matrix is: $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$.

On the one hand, we always use the multiplication by transformation matrix when we want to receive a matrix in a new basis:
- $A^{\prime }=CA$ , where $A^{\prime }$ is a matrix in a new basis; $C$ is a transformation matrix; $A$ is a matrix in the standard basis.
However, when we speak about getting the matrix in its eigenbasis, we use the other formula.
- $A=C^{-<!--\; -\; -->1}{A}^{\prime }C$ where $A$ is a matrix in a standard basis; $A^{\prime }$ is a matrix in the eigenbasis; $C$ is the transformation matrix.
Why do we have different formulas while doing the same thing (matrix transformations) or do I get something wrong?

For $x\in <!--\; \in \; -->\mathbb{C}$ define $A,B\in <!--\; \in \; -->M(3\times <!--\; \times \; -->3,\mathbb{C})$ as
$A=\left(\begin{array}{ccc}x& 0& 0\\ 0& x& 1\\ 0& 0& x\end{array}\right)$ and $B=\left(\begin{array}{ccc}x& 1& 0\\ 0& x& 0\\ 0& 0& x\end{array}\right).$
How to find a transformation matrix $T\in <!--\; \in \; -->{\text{GL}}_3(\mathbb{C})$ with $B=T^{-<!--\; -\; -->1}AT$

In linear algebra, to convert the linear transformation or operator into matrix form, it's not hard, get the standard basis of domain and substitute in the transformation and write the image of the standard basis as a linear combination of standard basis of codomain and finally get the constants (c i's) and write that in column wise in the matrix format, we will get the matrix format of the linear transformation with respect to the standard basis of the domain and codomain. My question here is "what about the reverse part?, can we find the linear transformation if the matrix is given?" I have tried multiplying the matrix with the standard basis of Mn(R), but I am getting the columns as the image.. What should I do? To get the transformation.

How can I solve the following exercise?
Let $V$ be the vector space defined on the field $\mathbb{R}$ of $n-<!--\; -\; -->$dimensional. Let $T:V\to <!--\; \to \; -->V$ be a linear transformation such that $T^{n-<!--\; -\; -->1}\ne <!--\; \ne \; -->0$, $T^n=0$ and $B=\{x,T\left(x\right),T^2\left(x\right),\cdots <!--\; \cdots \; -->,T^{n-<!--\; -\; -->1}\left(x\right)\}$ it is a basis for $V$. Find a matrix associated with the linear transformation $T$.
I think the only way to explicitly determine the matrix associated with the linear transformation $T$, is to define $T$ as
$T({\mathrm{\alpha <!--\; \alpha \; -->}}_{1}x+{\mathrm{\alpha <!--\; \alpha \; -->}}_{2}T\left(x\right)+\cdots <!--\; \cdots \; -->+{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}{T}^{n-<!--\; -\; -->1}\left(x\right))={\mathrm{\alpha <!--\; \alpha \; -->}}_{1}x+{\mathrm{\alpha <!--\; \alpha \; -->}}_{2}T\left(x\right)+\cdots <!--\; \cdots \; -->+{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}{T}^{n-<!--\; -\; -->1}\left(x\right)$.

I have a rigid 3D homogeneous transformation matrix 4*4 which includes Rotation (Roll, Yaw and pitch) plus the translation(X,Y and Z).
I want to split this transformation into N small and equal transformations where:
$T_{\text{Original}}=T.T.T.....T$
Is it theoretically OK to just extract the angles and translation values from $T_{\text{Original}}$ and divide them by $N$ then construct the $T$ matrix from them? If not, how can I achieve that?

$T(x,y)=(2x-<!--\; -\; -->y,3x-<!--\; -\; -->y)$
with respect to basis (1,0), and (0,1) Now M(T)=
$\left[\begin{array}{ccc}2& 3& x\\ -<!--\; -\; -->1& -<!--\; -\; -->1& y\end{array}\right]$
gives
$\left(\begin{array}{c}2x+3y\\ -<!--\; -\; -->x-<!--\; -\; -->y\end{array}\right)$
so this cannot be the transformation matrix
The correct one is
$\left[\begin{array}{ccc}2& -<!--\; -\; -->1& x\\ 3& -<!--\; -\; -->1& y\end{array}\right]$
Also, how do I find the matrix transformation of (x,y,z)=(x+y,y+z) for basis (1,0) and (0,1). So we are going from 3 dimensions to 2. So my matrix should have 2 rows and 2 columns right?

How to do this question in regards of matrices and transformation?

From my understanding, this question asks me is there any formula which can make $T(x)=x^2$ a non-linear transformation into a matrix transformation ? What am I supposed to answer to this question?

Assuming we have a transformation $T(x)=A\ast <!--\; \ast \; -->x$ what is the process for finding the inverse of the transformation $T^{-<!--\; -\; -->1}(x)$?
For example take
$A=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)$

Let $T$ be the transformation of 2 by 2 real symmetric matrices defined by:
$\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]$
$\left[\begin{array}{cc}c& -<!--\; -\; -->b\\ -<!--\; -\; -->b& a\end{array}\right]$
then which of the following statements is NOT true?
1. $det(T)=-<!--\; -\; -->1$
2. $T^{-<!--\; -\; -->1}=T$
3. $T$ is linear
4. the space of 2 by 2 real symmetric matricies with only zeros in the main diagonal is an eigenspace of $T$.
5. $\mathrm{\lambda <!--\; \lambda \; -->}=2$ is an eigen value of $T$

Acos 90 degree matrix transformation.
I'm writing a program that transforms a matrix of points by 90°. In it, I have two vectors from which I am performing the rotation. Both vectors are normalized:
$$\begin{gathered} A:x:\sqrt{(}\frac{1}{3}),y:\sqrt{(}\frac{1}{3}),z:-<!--\; -\; -->\sqrt{(}\frac{1}{3}) \\ B:x:\sqrt{(}\frac{1}{3}),y:\sqrt{(}\frac{1}{3}),z:\sqrt{(}\frac{1}{3}) \end{gathered}$$
As I visualize it, these two vectors are separated by 90°, but the dot product of these vectors comes out to $\frac{1}{3}$
$$\begin{gathered} \sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->\sqrt{(}\frac{1}{3})+\sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->\sqrt{(}\frac{1}{3})+\sqrt{(}\frac{1}{3})\ast <!--\; \ast \; -->-<!--\; -\; -->\sqrt{(}\frac{1}{3}) \\ =\frac{1}{3}+\frac{1}{3}-<!--\; -\; -->\frac{1}{3} \\ =\frac{1}{3} \end{gathered}$$
My code is then supposed to use arc-cos to come up with 90° from this number, but I believe arc-cos needs an input of 0 in order to produce a result of 90°. What am I missing here?