IMC / 1995 / Problems / Day 2, P7

IMC 1995 · Day 2 · P7

Let $A$ be $3 \times 3$ real matrix such that the vectors $Au$ and $u$ are orthogonal for each column vector $u \in \mathbb{R}^3$ . Prove that:

a) $A^\top = -A$ , where $A^\top$ denotes the transpose of the matrix $A$ ;

b) there exists a vector $v \in \mathbb{R}^3$ such that $Au = v \times u$ for every $u \in \mathbb{R}^3$ , where $v \times u$ denotes the vector product in $\mathbb{R}^3$ .

Solution (official)

a) Set $A = (a_{ij})$ , $u = (u_1, u_2, u_3)^\top$ . If we use the orthogonality condition $\tag{1} (Au, u) = 0$ with $u_i = \delta_{ik}$ we get $a_{kk} = 0$ . If we use (1) with $u_i = \delta_{ik} + \delta_{im}$ we get $a_{kk} + a_{km} + a_{mk} + a_{mm} = 0$ and hence $a_{km} = -a_{mk}$ .

b) Set $v_1 = -a_{23}$ , $v_2 = a_{13}$ , $v_3 = -a_{12}$ . Then $Au = (v_2 u_3 - v_3 u_2,\ v_3 u_1 - v_1 u_3,\ v_1 u_2 - v_2 u_1)^\top = v \times u.$

IMC 1995 · Day 2 · P7

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