Hey I have this exercise and I want to see if what I have done is correct.
Let $v_1$ and $v_2$ be a basis of the vector space $V$. Let $f: V \rightarrow V$ be the $K$-linear function with $f(v_1) = av_1 + bv_2$ and $f(v_2) = −bv_1 + av_2$. I must find the representing matrix for $id_V , f$ and $f^2$, if both the domain $V$ and the domain $V$ are based on the basis $v_1, v_2$.
We can find the representing matrix for $id_V$, $f$, and $f^2$ by computing the images of the basis vectors $v_1$ and $v_2$ under these linear transformations and representing them as column vectors in the matrix with respect to the given basis.
$id_V$ is the identity function, which maps every vector to itself. Therefore, the image of each basis vector under $id_V$ is simply the same vector. The representing matrix for $id_V$ is therefore the identity matrix with respect to the basis $v_1, v_2$: $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$f(v_1) = av_1 + bv_2$ and $f(v_2) = -bv_1 + av_2$. To find the representing matrix for $f$ with respect to the basis $v_1, v_2$, we compute the images of the basis vectors and represent them as column vectors: $f(v_1) = av_1 + bv_2 = \begin{pmatrix} a \\ b \end{pmatrix}$
$f(v_2) = -bv_1 + av_2 = \begin{pmatrix} -b \\ a \end{pmatrix}$
Therefore, the representing matrix for $f$ is:
$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$
To find the representing matrix for $f^2$, we first compute $f^2(v_1)$ and $f^2(v_2)$: $f^2(v_1) = f(f(v_1)) = f(av_1 + bv_2) = a f(v_1) + b f(v_2) = a\begin{pmatrix} a \\ b \end{pmatrix} + b\begin{pmatrix} -b \\ a \end{pmatrix} = \begin{pmatrix} a^2 - b^2 \\ 2ab \end{pmatrix}$
$f^2(v_2) = f(f(v_2)) = f(-bv_1 + av_2) = -b f(v_1) + a f(v_2) = -b\begin{pmatrix} a \\ b \end{pmatrix} + a\begin{pmatrix} -b \\ a \end{pmatrix} = \begin{pmatrix} -2ab \\ a^2 - b^2 \end{pmatrix}$
Therefore, the representing matrix for $f^2$ is:
$\begin{pmatrix} a^2 - b^2 & -2ab \\ 2ab & a^2 - b^2 \end{pmatrix}$
Am I right or am I doing something wrong?