# Relation between linear transformations and traces

a) Let $A \in M_n (K)$. We denote $f_A$ the linear form defined, for every $X \in M_n (K)$, by $f_A(X)=Tr(AX)$. Show that the function $f$ which maps $A \in M_n (K)$ to $f_A$ is an isomorphism between $M_n (K)$ and its dual.

b) Let $f: M_n (K) \rightarrow K$ be a linear form such that, for every $(X,Y)$ in $M_n (K)^2$, $f(XY)=f(YX)$. Show that there exists $\lambda \in K$ such that for every $X \in M_n (K)$, $f(X)=\lambda Tr(X)$

That is what I have:

a) Let $(E_{ij})_{1\leq i,j \leq n}$ be the standard basis for $M_n (K)$. Let us start by showing that for every $1 \leq i,j,k,l \leq n$, we have $E_{ij} E_{kl}= \delta_{jk} E_{il}$

We have $E_{ij}=(\delta_{pi} \delta_{qj})_{1 \leq p,q \leq n}$ and $E_{kl}=(\delta_{pk} \delta_{ql})_{1 \leq p,q \leq n}$ $A= E_{ij} E_{kl}= (a_{p,q})$ such that: $a_{p,q}= \sum \limits_{r=1}^n (\delta_{pi} \delta_{rj})(\delta_{rk} \delta_{ql})=( \sum \limits_{r=1}^n \delta_{rj} \delta_{rk})\delta_{pi} \delta_{ql}= \delta_{jk} \delta_{pi} \delta_{ql}$

Hence: $E_{ij} E_{kl}= \delta_{jk} E_{il}$ $$\begin{array}{c}\\\\\end{array}$$

It is obvious that $f$ is linear. Therefore, for dimensional reasons, we must only show that $f$ is injective. Let $A= (a_{ij})_{1\leq i,j \leq n}$ such that $f_A=0$. We therefore have, for $1 \leq i_0, j_0 \leq n$

$0= Tr(AE_{i_0 j_0})= Tr(\sum \limits_{1 \leq i,j \leq n} a_{ij}E_{ij}E{i_0 j_0})= Tr(\sum \limits_{i=1}^n a_{i i_{0}}E_{i i_0} E_{i_0 j_0})= \sum \limits_{i=1}^n a_{i i_0} Tr(E_{i j_0})=a_{j_{0} i_{0}}$

Hence $A$ equals zero.
Therefore, $f_A$ is an isomorphism. $$\begin{array}{c}\\\\\end{array}$$

Can someone help me with b?

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I am having trouble with question b – Carpediem Apr 6 '13 at 19:51

b) By assumption, $f$ is a linear form so by a) we have $A$ such that $f=f_A$.

Then for every $i\neq j$: $$a_{ji}=\mbox{Tr}(AE_{ij})=\mbox{Tr}(AE_{ii}E_{ij})=f_A(E_{ii}E_{ij})=f_A(E_{ij}E_{ii})=f_A(0)=0$$ and $$a_{ii}=\mbox{Tr}(AE_{ii})=f_A(E_{ii})=f_A(E_{ij}E_{ji})=f_A(E_{ji}E_{ij})=f_A(E_{jj})=\mbox{Tr}(AE_{jj})=a_{jj}.$$ Hence $A$ is a scalar matrix, which yields the result.

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Ok. What do you think of my answer below? I tried something.. – Carpediem Apr 6 '13 at 19:54
Motivation for the downvote? Just because the other answer is "better"? If so, it suffices to upvote the other answer. – 1015 Apr 6 '13 at 20:38
@julien I didn't downvote you – user43418 Apr 6 '13 at 23:01
@user43418 Ok. The mute downvoter will remain in the dark. – 1015 Apr 6 '13 at 23:08

Let $A \in M_n (K)$ such that $f=f_A$. We have, for every $(X,Y) \in M_n (K)^2$, $Tr(AXY)=Tr(AYX)$. Since $Tr(AYX)=Tr(XAY)$, we can deduce that $Tr((AX-XA)Y)=0$. Since this is true for every matrix $Y$, we have, according to question a), $AX=XA$. Therefore $A$ is commutative with every matrix $X$.

Let us show that $A$ is a scalar matrix i.e. a scalar multiple of the identity matrix. If $A= (a_{ij})_{1 \leq i,j \leq n}$, we have for every $1 \leq i,j \leq n$,

$AE_{ij}=\sum \limits_{1 \leq k,l \leq n} a_{kl} E_{kl} E_{ij}= \sum \limits_{k=1}^n a_{ki}E_{kj}=E_{ij}A= \sum \limits_{1 \leq k,l \leq n} a_{kl} E_{ij} E_{kl}= \sum \limits_{l=1}^n a_{jl} E_{il}$

By uniqueness of the writing, we obtain $a_{ki}=0$ for $k \neq i$ and $a_{ii}=a_{jj}$: A is therefore a scalar matrix. Hence, $f=f_A$ is collinear to the trace. Therefore, there exists $\lambda \in K$ such that for every $X \in M_n (K)$, $f(X)=\lambda Tr(X)$

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Yep, that's nice. – 1015 Apr 6 '13 at 20:01