Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 2

True. The symmetric and antisymmetric parts of a matrix are, in Einstein summation convention,They satisfy $S_{ji}=S_{ij}$, $A_{ji}=-A_{ij}$ and $S_{ij}+A_{ij}=M_{ij}$. This works over the real or complex numbers; it uses division by two.

False in general. For symmetric matrices, $(AB{)}_{ji}=A_{jk}{B}_{ki}=B_{ik}{A}_{kj}=(BA{)}_{ij}$. The symmetric product criterion for symmetric matrices is therefore $AB=BA$. For example,has $(AB{)}_{12}=1$ and $(AB{)}_{21}=0$ despite both factors being symmetric.

True. Set $C_{ij}=A_{ki}{B}_{k\ell }{A}_{\ell j}$. Exchange $i,j$, interchange the dummy indices $k,\ell$, and use $B_{\ell k}=B_{k\ell }$:Thus the matrix congruence $A^{T}BA$ preserves symmetry, even when $A$ is singular.

True. A symmetric tensor and an antisymmetric tensor are orthogonal under full index contraction. Renaming the two summed indices givesso $\boxed{A_{ij}{B}_{ij}=0}$. This is a scalar statement with both indices summed, not a claim that every individual entrywise product vanishes.

True. Orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $A_{ik}{A}_{jk}={\delta }_{ij}$, so $A^T=A^{-1}$. Thus $C_{ij}=A_{ki}{B}_{k\ell }{A}_{\ell j}$ is a similar matrix to $B$. If $B_{ij}{v}_j=\lambda v_i$, set $w_i=A_{ji}{v}_j$. ThenThe transformed eigenvector is nonzero because $A$ is invertible. Applying the inverse transformation gives the reverse correspondence, and similar matrices also have the same characteristic polynomial, including algebraic multiplicities of all eigenvalues.