Davon Irwin

2022-06-21

We have a linear transformation $T:{M}_{2×2}\left(F\right)\to F$ by $T\left(A\right)=tr\left(A\right)$.
We want to compute the matrix representation $\left[T\right]$ from $\alpha$ to $\gamma$ coordinates.
${M}_{2×2}$ has the standard ordered basis "$\alpha$" for a $2×2$ matrix. $F$ has the standard basis "$\gamma$" for a scalar.
My understanding is that any matrix $A$ in the space ${M}_{2×2}$ can be represented by the linear combination:
$a{\alpha }_{1}+b{\alpha }_{2}+c{\alpha }_{3}+d{\alpha }_{4}$
and the $tr\left(A\right)$ can be written as:
$\left(a+c\right)\gamma .$
I'm not sure how to get the matrix representation from this.

### Answer & Explanation

podesect

Our linear map is
$\begin{array}{ccccc}T& :& {\mathcal{M}}_{2×2}\left(F\right)& \to & F\\ & & A& ↦& \mathrm{tr}\left(A\right)\end{array}$
The basis $\alpha$ for ${\mathcal{M}}_{2×2}\left(F\right)$ is
$\alpha =\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right\}$
and the basis $\gamma$ for $F$ is $\gamma =\left\{1\right\}$.
Now, note that
$\begin{array}{lcr}T\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)& =& 1\\ T\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)& =& 0\\ T\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)& =& 0\\ T\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)& =& 1\end{array}$
This implies that $\left[T{\right]}_{\alpha }^{\gamma }$ is the $1×4$ matrix
$\left[T{\right]}_{\alpha }^{\gamma }=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]$
This allows us to interpret trace as matrix multiplication. Note that
$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=a\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+b\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]+c\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]+d\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$
which means that relative to $\alpha$ the matrix $A$ can be viewed as the vector
$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]$
Hence
$\mathrm{tr}\left(A\right)=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=a+d$

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