Hey I want to check my solutions with this problem. Can someone help me?
Let $K$ be a field, $n ∈ \mathbb{N}$, and $l: Mat_{n,n}(K) → K$ a linear form such that $l(AB) = l(BA)$ for all $A, B ∈ Mat_{n,n}(K)$.
Show that there exists a $λ ∈ K$ such that $l(A) = λtrace(A)$ for all $A ∈ Mat_{n,n}(K)$.
Remark: In particular, show that there is, up to scalar multiples, only one linear form $l: Mat_{n,n}(K) → K$ such that $l(AB) = l(BA)$ for all $A, B$.
To solve this problem, we first define the linear form $l: Mat_{n,n}(K) → K$ as $l(A) = trace(A)$ for all $A ∈ Mat_{n,n}(K)$, where$ trace(A)$ denotes the trace (i.e., the sum of the diagonal entries) of the matrix A.
Next, we show that this linear form satisfies the condition $l(AB) = l(BA)$ for all $A, B ∈ Mat_{n,n}(K)$:
$l(AB) = trace(AB) = trace(BA) = l(BA)$
Therefore, $l$ satisfies the given condition.
To show that any linear form satisfying the condition must be a scalar multiple of trace, let $l: Mat_{n,n}(K) → K$ be a linear form such that $l(AB) = l(BA)$ for all $A, B ∈ Mat_{n,n}(K)$. Let $E_{i,j}$ be the matrix with $1$ in the $(i,j)$ position and $0$ elsewhere, and let $e_i$ be the column vector with $1$ in the i-th position and $0$ elsewhere.
Then, for any $i$ and $j$, we have:
$l(E_{i,i}E_{j,j}) = l(E_{j,j}E_{i,i})$
$l(e_i^Te_j) = l(e_j^Te_i)$
Therefore, $l(e_i^Te_j) = 0$ for all $i ≠ j$.
Now, let $A = (a_{i,j})$ be any matrix in $Mat_{n,n}(K)$, and let $B = (b_{i,j})$ be the matrix defined by $b_{i,j} = e_i^Te_j$. Then, for any $i$ and $j$, we have:
$l(a_{i,j}b_{i,j}) = l(b_{i,j}a_{i,j})$
$a_{i,j}l(b_{i,j}) = b_{i,j}l(a_{i,j})$
$a_{i,j}l(e_i^Te_j) = e_i^Te_jl(a_{i,j})$
$a_{i,j}l(e_i)^T e_j = e_i^T l(A)e_j$
Therefore, $l(A) = λtrace(A)$ for some $λ ∈ K$, where $λ = l(e_i)^T$ for any $i$.
Hence, we have shown that any linear form satisfying the condition must be a scalar multiple of trace, and that the linear form $l(A) = trace(A)$ satisfies the condition. Therefore, there exists a unique (up to scalar multiples) linear form $l: Mat_{n,n}(K) → K$ such that $l(AB) = l(BA)$ for all $A, B ∈ Mat_{n,n}(K)$, and it is given by $l(A) = λtrace(A)$ for some $λ ∈ K$.
At first glance, the general idea looks fine. We can clean things up slightly.
A linear form is determined by its values on a basis, so it suffices to show that $\ell(E_{ij})=0$ when $i\neq j$ and $\ell(E_{ii})=\ell(E_{jj})$. Using the multiplication relation between the one-non-zero element matrices $E_{ab}E_{cd}=\delta_{bc}E_{ad}$, we get rather clean computations.
For the first claim $$\ell(E_{ij})=\ell(E_{ij}E_{jj})=\ell(E_{jj}E_{ij})=\ell(0)=0$$
For the second claim,
$$\ell(E_{ii})=\ell(E_{ij}E_{jj}E_{ji})=\ell(E_{jj}E_{ji}E_{ij})=\ell(E_{jj}^2)=\ell(E_{jj}).$$
To be clear, this is just a slightly cleaner reformulation of what I believe your underlying ideas/computations are.