Let $V$ be the space of all $2\times 2$ matrices over the field $F$ and let $P$ be a fixed $2\times 2$ matrix over $F$. Let $T$ be the linear operator on $V$ defined by $T(A)=PA$, for ever $A$ belonging to $V$ . Prove that $tr(T)= 2 tr(P)$.
Also what all are the formula's constituting trace of a matrix?
On
Hint: $T$ is a map from the 4 dimensional space $V$ of 2 by 2 matrices to itself. In coordinates you may write this as: $$ (T(A))_{ij} = \sum_k P_{ik}A_{kj} = \sum_k\sum_j [P_{ik}\delta_{jl}] A_{kl}$$ So you should calculate the trace of $M_{ij,kl} = P_{ik}\delta_{jl}$ which is $$ {\rm tr} \; M = \sum_{ij} M_{ij,ij}= \sum_{i=1}^2\sum_{j=1}^2 P_{ii} \delta_{jj}= 2\; {\rm tr}\; P $$
Let $P:=\pmatrix{a&c\\b&d}$ with tr($P$)=$a+d.$ Let us consider operator $T$ defined by:
$$\tag{0}T(A)=PA$$
First solution:
Setting $A:=\pmatrix{x&z\\y&t}$ and $PA:=\pmatrix{x'&z'\\y'&t'}$, formula (0) is equivalent to:
$$\begin{cases}x'&=&ax+cy\\y'&=&bx+dy\\z'&=&az+ct\\t'&=&bz+dt\end{cases}$$
Otherwise said, to:
$$\tag{*}\pmatrix{x'\\y'\\z'\\t'}=\pmatrix{a&c&0&0\\b&d&0&0\\0&0&a&c\\0&0&b&d}\pmatrix{x\\y\\z\\t}$$
The trace of this matrix is clearly twice the trace of $P$.
Remark: The trace of operator $T$ has been computed with a certain matrix, but it is the same with respect to any other since it is an invariant.
Second solution (that has strong connections with the previous solution):
Consider the canonical basis on the vector space of $2 \times 2$ matrices $\frak{M_2}$ with entries in field $F$:
$$E_1=\pmatrix{1&0\\0&0}, \ E_2=\pmatrix{0&0\\1&0}, \ E_3=\pmatrix{0&1\\0&0}, \ E_4=\pmatrix{0&0\\0&1}.$$
Let us compute the images of the $E_k$s by operator T:
$$T(E_1)=\pmatrix{a&c\\b&d}\pmatrix{1&0\\0&0}=\pmatrix{a&0\\b&0}=aE_1+bE_3.$$
In a similar way:
$$T(E_2)=\pmatrix{c&0\\d&0}=cE_1+dE_2$$
$$T(E_3)=\pmatrix{0&a\\0&b}=aE_3+bE_4$$
$$T(E_4)=\pmatrix{0&c\\0&d}=cE_3+dE_4.$$
Thus the matrix of operator $T$ with respect to the canonical basis is:
$$\tag{1}[T]=\pmatrix{a&c&0&0\\b&d&0&0\\0&0&a&c\\0&0&b&d}$$
as obtained in the first solution, with the same conclusion.
Third (shorter) solution: The RHS of (0) can be transformed under the following form, using concepts and results on Kronecker product that can be found in this reference:
$$\tag{2}PA=PAI_2=(I_2^T \otimes P)vec(A)=(I_2 \otimes P)vec(A)$$ where vec$(A)$ is obtained by piling up the two colums of $A$.
Formula (2) clearly amounts to formula (*).
An important thing is that, with this method, we have not to give an explicit form to the matrix of transformation $T$; it suffices to use the general formula: tr$(A \otimes B) =$ tr$(A) $tr$(B)$ (see the same Wikipedia reference).
Were the same question been asked for $n \times n$ matrices, we would have at once the answer $n \ $tr$(P)$.