# Determine basis of matrix

I have this example and by now I am able to do only a) and maybe b). I would be thankful is someone could help me with the rest.

a) Determine rotation matrix with angle $\pi$

What I did:

$$D=\begin{pmatrix} \cos\pi & -\sin\pi \\ \sin\pi & \cos\pi \end{pmatrix}$$

b) Determine basis B that emerge from basis E2 through roation with angle $\pi$.

Here I think I should just multiply rotation matrix with $E_{2}=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$.
And then I will get the same matrix. Is this correct?

c) Determine matrix A with linear mapping $\phi: \mathbb R ^{2}\mapsto \mathbb R ^{2}$ which is given through

$$\phi \left(x_{1},x_{2} \right)=\begin{pmatrix} 5x_{1}+3x_{2} \\ x_{2} \end{pmatrix}$$

Here I have no idea how to start.

d) Determine coordinates of $\vec{v}$ and $\phi\left(\vec{v} \right)$ with respect to basis B.

To get coordinates of $\vec{v}$, maybe to multiply this vector with inverse matrix from b)? For $\phi\left(\vec{v} \right)$ I dont know.

For $(b)$, Notice that $D$ is an invertible matrix, so you are just rotating both basis vectors to get two new basis vectors $\begin{pmatrix}-1\\0\end{pmatrix},\begin{pmatrix}0\\-1\end{pmatrix}$. Notice that the length of those new basis vectors are the same, that is because $D$ is an "orthogonal," or distance preserving, matrix.
$(c)$ For any linear transformation $T:A\to B$, with bases $a_1,\ldots,a_n$ and $b_1,\ldots b_m$, and maps $\alpha:A\to F^n$ and $\beta:B\to F^m$ given by $$\alpha(\sum \lambda_ia_i)=\begin{pmatrix}\lambda_1\\\vdots\\\lambda_n\end{pmatrix}$$ and $$\beta(\sum \lambda_ib_i)=\begin{pmatrix}\lambda_1\\\vdots\\\lambda_m\end{pmatrix},$$
we take the matrix of the linear transformation $T$ to be given by $$M=\begin{pmatrix}\vdots&&\ldots&&\vdots\\ \beta(T(a_1))&&\ldots&&\beta(T(a_n))\\\vdots&&\ldots&&\vdots\end{pmatrix}.$$ This is because $$M\alpha(a_i)=M\cdot\begin{pmatrix}0\\\vdots\\1\\\vdots\\0\end{pmatrix}=\beta(T(a_i)),$$ which is the defining relation of a matrix. I now leave it to you to figure out $(c)$.
$(d)$ These are just the maps $\alpha$ and $\beta$-- just write down $\vec v$ and $\phi(\vec v)$ as linear combinations of the basis and the vectors are the components.
For example, with the basis $\begin{pmatrix}-1\\0\end{pmatrix},\begin{pmatrix}0\\-1\end{pmatrix}$, and the vector $v=\begin{pmatrix}1\\-2\end{pmatrix}=-1\begin{pmatrix}-1\\0\end{pmatrix}+2\begin{pmatrix}0\\-1\end{pmatrix}$ corresponds to the coordinate vector $\begin{pmatrix}-1\\2\end{pmatrix}$, and so $\alpha(\begin{pmatrix}1\\-2\end{pmatrix})=\begin{pmatrix}-1\\2\end{pmatrix}$. In your case $\beta=\alpha$ because $A=B=\mathbb R^2$ and you are using the same basis for both (I presume).
It is important to make the distinction between the vector space $V$ and the coordinate vector space $F^n$, except in you case they are both $\mathbb R^2$. I hope this is not too much of a mouthfull, but I promise as soon as you understand this relation, linear algebra will be easy.