IMC / 1997 / Problems / Day 2, P10

IMC 1997 · Day 2 · P10

a) Let the mapping $f : M_n \to \mathbb{R}$ from the space $M_n = \mathbb{R}^{n^2}$ of $n \times n$ matrices with real entries to reals be linear, i.e.: $\tag{1} f(A + B) = f(A) + f(B), \quad f(cA) = c f(A)$ for any $A, B \in M_n$ , $c \in \mathbb{R}$ . Prove that there exists a unique matrix $C \in M_n$ such that $f(A) = \operatorname{tr}(AC)$ for any $A \in M_n$ . (If $A = \{a_{ij}\}_{i,j=1}^{n}$ then $\operatorname{tr}(A) = \sum\limits_{i=1}^{n} a_{ii}$ ).

b) Suppose in addition to (1) that $\tag{2} f(A \cdot B) = f(B \cdot A)$ for any $A, B \in M_n$ . Prove that there exists $\lambda \in \mathbb{R}$ such that $f(A) = \lambda \cdot \operatorname{tr}(A)$ .

Solution (official)

a) If we denote by $E_{ij}$ the standard basis of $M_n$ consisting of elementary matrix (with entry 1 at the place $(i,j)$ and zero elsewhere), then the entries $c_{ij}$ of $C$ can be defined by $c_{ij} = f(E_{ji})$ .

b) Denote by $L$ the $n^2 - 1$ -dimensional linear subspace of $M_n$ consisting of all matrices with zero trace. The elements $E_{ij}$ with $i \ne j$ and the elements $E_{ii} - E_{nn}$ , $i = 1, \dots, n-1$ form a linear basis for $L$ . Since $\begin{align*} E_{ij} &= E_{ij} \cdot E_{jj} - E_{jj} \cdot E_{ij}, \quad i \ne j \\ E_{ii} - E_{nn} &= E_{in} \cdot E_{ni} - E_{ni} \cdot E_{in}, \quad i = 1, \dots, n-1, \end{align*}$ then the property (2) shows that $f$ is vanishing identically on $L$ . Now, for any $A \in M_n$ we have $A - \frac{1}{n} \operatorname{tr}(A) \cdot E \in L$ , where $E$ is the identity matrix, and therefore $f(A) = \frac{1}{n} f(E) \cdot \operatorname{tr}(A)$ .

IMC 1997 · Day 2 · P10

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