Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 18

The matrix trace is the sum of diagonal entries:ConsequentlyThe first diagonal entry is $a^2$ in the original PDF.

Expanding the determinant along the first row givesUnder the matrix trace condition $a^2=3$, this simplifies to $\det M=b(3-b)$. A square matrix is invertible exactly when its determinant is nonzero. Thus

The characteristic polynomial of $A$ isIts distinct eigenvalues are $3$ and $-3$. For $3$, the eigenvector relation gives $y=-2x$, so choose $(1,-2{)}^T$. For $-3$, it gives $x=-2y$, so choose $(2,-1{)}^T$. Put these independent eigenvectors into columns:Then $AB=BC$ column by column. Also $\det B=3\ne 0$, so this choice supplies the diagonalization of a matrix $B^{-1}AB=C$. Scaling columns or interchanging their order gives other valid choices, with the matching order in $C$.

For the choice of $B$ in part (b)(i), the two-by-two matrix inverse formula givesMultiplication gives $B{B}^{-1}=B^{-1}B=I$, providing a direct check.

The expression assumes $Y$ is invertible. By the cyclic property of the trace,For a real or complex antisymmetric matrix, $X^T=-X$ forces every diagonal entry to satisfy $x_{ii}=-x_{ii}$, hence $x_{ii}=0$. ThereforeThis also demonstrates invariance of the matrix trace under matrix similarity. Antisymmetry of $Z$ itself is not needed.

An orthogonal matrix has orthonormal columns. The first column of $L$ has squared norm $1+c^2$, so $c=0$. Orthogonality of the two columns then gives $-b+cd=-b=0$, hence $b=0$. The second column must have squared norm $d^2=1$. ThereforeBoth resulting matrices satisfy $L^{T}L=I$ directly.

Among the two orthogonal matrices from part (d)(i), $d=-1$ gives $L=-I$, a rotation matrix through angle $\pi$. For $d=1$, the determinant is $-1$, so the matrix reverses orientation and is a reflection rather than a rotation:It is the reflection matrix across the $y$ axis.

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.