Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 17

Suppose $B\mathbf{x}=0$. The given quadratic form bound then givesso $\mathbf{x}=0$. Thus the kernel of $B$ is trivial. A square matrix with trivial kernel is an invertible matrix, proving the assertion. No symmetry assumption on $B$ is needed.

The two prescribed directions have dot product zero. Normalize them toTheir cross product is the perpendicular unit vector ${\mathbf{q}}_3=(2,-1,-1{)}^T/\sqrt{6}$. Using these as columns givesThe columns form an orthonormal basis, so $Q^{T}Q=I$ and $Q$ is an orthogonal matrix.

The determinant of $P$ is $2\cdot 1-1(-1)=3$. The two-by-two inverse matrix formula yields

Direct matrix multiplication givesThus $D$ is diagonal. The columns of $P$ are eigenvectors of $A$ with eigenvalues one and four.

The similarity transformation gives $A=PD{P}^{-1}$. In the matrix power, each adjacent pair $P^{-1}P$ cancels to the identity matrix:This also follows by mathematical induction: the formula holds for $n=1$, and multiplication by $A=PD{P}^{-1}$ turns its $n$th instance into the $(n+1)$st.

Since $D^6=\mathrm{diag}(1,4^6)=\mathrm{diag}(1,4096)$,