Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 8

The trace sums the diagonal entries, so $\mathrm{tr}P={\sum }_{i}1-{\sum }_i{a}_i^2-{\sum }_i{b}_i^2$. The two unit vectors have squared Euclidean norms one, giving $\boxed{\mathrm{tr}P=n-2}$.

Write $P=I-Q$ with $Q=a{a}^T+b{b}^T$. The orthonormality conditions $a^{T}a=b^{T}b=1$ and $a^{T}b=b^{T}a=0$ giveConsequently $P^2=I-2Q+Q^2=I-Q=P$. It is the orthogonal projection matrix onto the orthogonal complement of the linear span of $a,b$.

The matrix product is $Pa=a-a(a^{T}a)-b(b^{T}a)=a-a=0$. Thus $\boxed{Pa=0}$, and the nonzero vector $a$ is in its kernel. Hence $P$ is singular and $\boxed{\det P=0}$. This includes $n=2$, when $a,b$ span the whole space and $P=0$.

The transpose reverses a matrix product. Since $S^T=S$, we have $(S^2{)}^T=S^T{S}^T=S^2$. Therefore $S^2$ is a symmetric matrix.

Apply trace orthogonality of symmetric and antisymmetric matrices. In detail, invariance of the trace under transpose and its cyclic property giveTherefore $\boxed{\mathrm{tr}(SA)=0}$. The negative sign comes from $A^T=-A$, while $S^T=S$.