Davon Irwin

2022-06-21

We have a linear transformation $T:{M}_{2\times 2}(F)\to F$ by $T(A)=tr(A)$.

We want to compute the matrix representation $[T]$ from $\alpha $ to $\gamma $ coordinates.

${M}_{2\times 2}$ has the standard ordered basis "$\alpha $" for a $2\times 2$ matrix. $F$ has the standard basis "$\gamma $" for a scalar.

My understanding is that any matrix $A$ in the space ${M}_{2\times 2}$ can be represented by the linear combination:

$a{\alpha}_{1}+b{\alpha}_{2}+c{\alpha}_{3}+d{\alpha}_{4}$

and the $tr(A)$ can be written as:

$(a+c)\gamma .$

I'm not sure how to get the matrix representation from this.

We want to compute the matrix representation $[T]$ from $\alpha $ to $\gamma $ coordinates.

${M}_{2\times 2}$ has the standard ordered basis "$\alpha $" for a $2\times 2$ matrix. $F$ has the standard basis "$\gamma $" for a scalar.

My understanding is that any matrix $A$ in the space ${M}_{2\times 2}$ can be represented by the linear combination:

$a{\alpha}_{1}+b{\alpha}_{2}+c{\alpha}_{3}+d{\alpha}_{4}$

and the $tr(A)$ can be written as:

$(a+c)\gamma .$

I'm not sure how to get the matrix representation from this.

podesect

Beginner2022-06-22Added 20 answers

Our linear map is

$\begin{array}{ccccc}T& :& {\mathcal{M}}_{2\times 2}(F)& \to & F\\ & & A& \mapsto & \mathrm{tr}(A)\end{array}$

The basis $\alpha $ for ${\mathcal{M}}_{2\times 2}(F)$ is

$\alpha =\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\}$

and the basis $\gamma $ for $F$ is $\gamma =\{1\}$.

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 $[T{]}_{\alpha}^{\gamma}$ is the $1\times 4$ matrix

$[T{]}_{\alpha}^{\gamma}=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]$

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

$\mathrm{tr}(A)=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=a+d$

$\begin{array}{ccccc}T& :& {\mathcal{M}}_{2\times 2}(F)& \to & F\\ & & A& \mapsto & \mathrm{tr}(A)\end{array}$

The basis $\alpha $ for ${\mathcal{M}}_{2\times 2}(F)$ is

$\alpha =\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\}$

and the basis $\gamma $ for $F$ is $\gamma =\{1\}$.

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 $[T{]}_{\alpha}^{\gamma}$ is the $1\times 4$ matrix

$[T{]}_{\alpha}^{\gamma}=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]$

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

$\mathrm{tr}(A)=\left[\begin{array}{cccc}1& 0& 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=a+d$